PhysicalScience Gr10 CAPS

508
Physical Sciences Grade 10 WebBook Authors: Siyavula / Free High School Science Texts ISBN: 978-1-920423-08-7

Transcript of PhysicalScience Gr10 CAPS

Page 1: PhysicalScience Gr10 CAPS

Physical Sciences Grade 10 WebBook

Authors:

Siyavula / Free High School Science Texts

ISBN: 978-1-920423-08-7

Page 2: PhysicalScience Gr10 CAPS

This selection and arrangement of content as a collection is copyrighted by Siyavula / Free High School Science Texts. It is

licensed under the Creative Commons Attribution 3.0 license (http://creativecommons.org/licenses/by/3.0/).

Collection structure revised: June 13, 2011

Page 3: PhysicalScience Gr10 CAPS

FHSST Core TeamMark Horner ; Samuel Halliday ; Sarah Blyth ; Rory Adams ; Spencer Wheaton

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iii

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1 2.2

H

1,01

3 0.98

Li

6,94

11 0.93

Na

23,0

19 0.93

K

39,1

37 0.82

Rb

85,5

55 0.79

Cs

132,9

87 0.7

Fr

(223)

4 1.57

Be

9,01

12 1.31

Mg

24,3

20 1.00

Ca

40,1

38 0.95

Sr

87,6

56 0.89

Ba

137,3

88 0.9

Ra

226,0

21 1.36

Sc

45,0

39 1.22

Y

88,9

57-71

La-Lu

Lanthanides

89-103

Ac-Lr

Actinides

22 1.54

Ti

47,9

40 1.33

Zr

91,2

72 1.30

Hf

178,5

104

Rf

(261)

23 1.63

V

50,9

41 1.60

Nb

92,9

73 1.50

Ta

180,9

105

Db

(262)

24 1.66

Cr

52,0

42 2.16

Mo

95,9

74 2.36

W

183,8

106

Sg

(263)

25 1.55

Mn

54,9

43 1.90

Tc

(98)

75 1.90

Re

186,2

107

Bh

(262)

26 1.33

Fe

55,8

44 2.20

Ru

101,1

76 2.20

Os

190,2

108

Hs

(265)

27 1.88

Co

58,9

45 2.28

Rh

102,9

77 2.20

Ir

192,2

109

Mt

(266)

28 1.91

Ni

58,7

46 2.20

Pd

106,4

78 2.28

Pt

195,1

110

Ds

(269)

29 1.90

Cu

63,5

47 1.93

Ag

107,9

79 2.54

Au

197,0

111

Rg

(272)

30 1.65

Zn

65,4

48 1.69

Cd

112,4

80 2.00

Hg

200,6

112

Cn

(277)

31 1.81

Ga

69,7

13 1.61

Al

27,0

5 2.04

B

10,8

49 1.78

In

114,8

81 1.62

Tl

204,4

113

Uut

(284)

6 2.55

C

12,0

14 1.90

Si

28,1

32 2.01

Ge

72,6

50 1.96

Sn

118,7

82 2.33

Pb

207,2

114

Uuq

(289)

7 3.04

N

14,0

15 2.19

P

31,0

33 2.18

As

74,9

51 2.05

Sb

121,8

83 2.02

Bi

209,0

115

Uup

(288)

8 3.44

O

16,0

16 2.58

S

32,1

34 2.55

Se

79,0

52 2.10

Te

127,6

84 2.00

Po

(209)

116

Uuh

(293)

9 3.98

F

19,0

17 3.16

Cl

35,45

35 2.96

Br

79,9

53 2.66

I

126,9

85 2.20

At

(210)

117

Uus

(282)

10

Ne

20,2

2

He

4,00

18

Ar

39,9

36

Kr

83,8

54

Xe

131,3

86

Rn

(222)

118

Uuo

(282)

1 IA

2 IIA

3 IIIB 4 IVB 5 VB 6 VIB 7 VIIB 8 VII 9 VII 10 VII 11 IB 12 IIB

13 IIIA 14 IVA 15 VA 16 VIA 17 VIIA

18 0

57 1.10

La

138,9

58 1.12

Ce

140,1

59 1.13

Pr

140,9

60 1.14

Nd

144,2

61

Pm

(145)

62 1.17

Sm

150,4

63

Eu

152,0

64 1.20

Gd

157,3

65

Tb

158,9

66 1.22

Dy

162,5

67 1.23

Ho

164,9

68 1.24

Er

167,3

69 1.25

Tm

168,9

70

Yb

173,0

71 1.27

Lu

175,0

89 1.10

Ac

227,0

90 1.30

Th

232,0

91 1.50

Pa

231,0

92 1.38

U

238,0

93 1.36

Np

237,0

94 1.28

Pu

(244)

95 1.30

Am

(243)

96 1.30

Cm

(247)

97 1.30

Bk

(247)

98 1.30

Cf

(251)

99 1.30

Es

(252)

100 1.30

Fm

(257)

101 1.30

Md

(258)

102 1.30

No

(258)

103

Lr

(260)

Transition Metal

Metal

Metalloid

Non-metal

Noble Gas

Lanthanide

Actinide

Periodic Table of the Elements

No EN

Element

AMU

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Contents

1 Units 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Unit Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2.1 SI Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2.2 The Other Systems of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.3 c.g.s. Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.4 Imperial Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.5 Natural Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Writing Units as Words or Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.3.1 Naming of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.4 Combinations of SI Base Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.5 Rounding, Scientific Notation and Significant Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.5.1 Rounding Off . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.5.2 Error Margins . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.5.3 Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.5.4 Significant Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.5.5 Using Significant Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.6 Prefixes of Base Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.6.1 Using Scientific Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.7 The Importance of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.7.1 Investigation : Importance of Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.7.2 Case Study : The importance of units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.8 How to Change Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

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1.8.1 Two other useful conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.9 A sanity test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.9.1 The scale of the matter... : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.11 End of Chapter Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

I Chemistry 15

Classification of Matter 17

1.12 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.13 Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.13.1 Heterogeneous mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.13.2 Homogeneous mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.13.3 Mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.14 Pure Substances: Elements and Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.14.1 Activity: Smartie Chromatography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.14.2 Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1.14.3 Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Elements, mixtures and compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.15 Giving names and formulae to substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.15.1 Naming compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.16 Metals, Metalloids and Non-metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.16.1 Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Group Work : Looking at metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

1.16.2 Non-metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.16.3 Metalloids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.17 Electrical conductors, semi-conductors and insulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

1.17.1 Experiment : Electrical conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

1.18 Thermal Conductors and Insulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.18.1 Demonstration : Thermal conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.18.2 Investigation : A closer look at thermal conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

1.19 Magnetic and Non-magnetic Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

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1.19.1 Investigation : Magnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.19.2 Group Discussion : Properties of materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.20 Seperating mixtures - Not in CAPS - included for completeness . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.20.1 Investigation : The separation of a salt solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

1.20.2 Discussion : Separating mixtures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.21 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.21.1 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

States of matter and the kinetic molecular theory 39

1.22 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

1.23 States of matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

1.23.1 Experiment: States of matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

1.24 The Kinetic Theory of Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

1.25 Intramolecular and intermolecular forces (Not in CAPS - Included for Completeness) . . . . . . . . . . . . . . 44

1.25.1 Intramolecular and intermolecular forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1.26 The Properties of Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

1.26.1 Exercise: Forces and boiling point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

1.26.2 Investigation : Determining the density of liquids: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1.26.3 Investigation : Determining the density of irregular solids: . . . . . . . . . . . . . . . . . . . . . . . . 49

1.27 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

1.28 End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

The Atom 53

1.29 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

1.30 Project: Models of the atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

1.31 Models of the Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

1.31.1 The Plum Pudding Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

1.31.2 Rutherford’s model of the atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

1.31.3 The Bohr Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

1.31.4 Other models of the atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1.31.5 Models of the atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1.32 The size of atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

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1.32.1 How heavy is an atom? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

1.32.2 How big is an atom? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

1.33 Atomic structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

1.33.1 The Electron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

1.33.2 The Nucleus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

The Proton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

The Neutron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

1.34 Atomic number and atomic mass number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

1.34.1 The structure of the atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

1.35 Isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

1.35.1 What is an isotope? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

Isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

1.35.2 Relative atomic mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Isotopes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

Group Discussion : The changing nature of scientific knowledge . . . . . . . . . . . . . . . . . . . . . 68

1.36 Electron configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

1.36.1 The energy of electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

1.36.2 Energy quantisation and line emission spectra (Not in CAPS, included for completeness) . . . . . . . . 69

1.36.3 Electron configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

1.36.4 Core and valence electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

1.36.5 The importance of understanding electron configuration . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Experiment: Flame tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Energy diagrams and electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Group work : Building a model of an atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

1.37 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

1.37.1 End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

The Periodic Table 83

1.38 The arrangement of atoms in the periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

1.38.1 Activity: Inventing the periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

1.38.2 Groups in the periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Investigation : The properties of elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

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1.38.3 Periods in the periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Exercise: Elements in the periodic table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

1.39 Ionisation Energy and the Periodic Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

1.39.1 Ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

Investigation : The formation of ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Exercise: The formation of ions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

1.39.2 Ionisation Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

Trends in ionisation energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

1.40 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

1.41 End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

Chemical Bonding 93

1.42 Chemical Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

1.43 What happens when atoms bond? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

1.44 Covalent Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

1.44.1 The nature of the covalent bond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

Covalent bonding and valency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

1.44.2 Properties of covalent compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

1.45 Lewis notation and molecular structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

1.45.1 Atomic bonding and Lewis notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

1.46 Ionic Bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

1.46.1 The nature of the ionic bond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

Ionic compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

1.46.2 The crystal lattice structure of ionic compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

1.46.3 Properties of Ionic Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

1.47 Metallic bonds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

1.47.1 The nature of the metallic bond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

1.47.2 The properties of metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

1.47.3 Activity: Building models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

1.47.4 Chemical bonding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

1.48 Writing chemical formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

1.48.1 The formulae of covalent compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

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1.48.2 The formulae of ionic compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

Chemical formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

1.49 Chemical compounds: names and masses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

1.50 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

1.50.1 End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

What are the objects around us made of 115

1.51 Introduction: The atom as the building block of matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

1.52 Compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

1.52.1 Representing molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

Atoms and molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

1.53 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

1.53.1 End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

Physical and Chemical Change 123

1.54 Physical and Chemical Change - Grade 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

1.55 Physical changes in matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

1.56 Chemical Changes in Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

1.56.1 Decomposition reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

Experiment : The decomposition of hydrogen peroxide . . . . . . . . . . . . . . . . . . . . . . . . . . 126

1.56.2 Synthesis reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Experiment: Chemical reactions involving iron and sulphur . . . . . . . . . . . . . . . . . . . . . . . . 128

Investigation : Physical or chemical change? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

1.57 Energy changes in chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

1.58 Conservation of atoms and mass in reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

1.58.1 Activity : The conservation of atoms in chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . 132

1.58.2 Experiment: Conservation of matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

1.59 Law of constant composition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

1.60 Volume relationships in gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

1.61 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

1.61.1 End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

Representing Chemical Change 137

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1.62 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

1.63 Chemical symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

1.63.1 Revising common chemical symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

1.64 Writing chemical formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

1.65 Balancing chemical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

1.65.1 The law of conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

1.65.2 Activity: Balancing chemical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

1.65.3 Steps to balance a chemical equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

Balancing simple chemical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

1.66 State symbols and other information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

1.66.1 Balancing more advanced chemical equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

1.67 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

1.67.1 End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

Reactions in Aqueous Solutions 151

1.68 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

1.69 Ions in aqueous solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

1.69.1 Dissociation in water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

Ions in solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

1.69.2 Ions and water hardness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

1.70 Acid rain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

1.70.1 Investigation : Acid rain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

1.71 Electrolytes, ionisation and conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

1.71.1 Electrolytes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

1.71.2 Non-electrolytes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

1.71.3 Factors that affect the conductivity of water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

Experiment : Electrical conductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

1.72 Types of reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

1.72.1 Precipitation reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

Demonstration : The reaction of ions in solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

1.72.2 Testing for common anions in solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

Test for a chloride . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

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Test for a sulphate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

Test for a carbonate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

Test for bromides and iodides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

Precipitation reactions and ions in solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

1.72.3 Other reactions in aqueous solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

Acid-base reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

Redox reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

Demonstration: Oxidation of sodium metal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

1.72.4 Experiment: Reaction types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

1.73 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

1.74 End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

Quantitative Aspects of Chemical Change 171

1.75 Quantitative Aspects of Chemical Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

1.76 The Mole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

1.76.1 Moles and mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

1.77 Molar Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

1.77.1 Moles and molar mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175

1.78 An equation to calculate moles and mass in chemical reactions . . . . . . . . . . . . . . . . . . . . . . . . . . 176

1.78.1 Some simple calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

1.79 Molecules and compounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178

1.79.1 Group work : Understanding moles, molecules and Avogadro’s number . . . . . . . . . . . . . . . . . 180

1.79.2 More advanced calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

1.80 The Composition of Substances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

1.80.1 Moles and empirical formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

1.81 Molar Volumes of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

1.81.1 Using the combined gas law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187

1.82 Molar concentrations of liquids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

1.82.1 Molarity and the concentration of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

1.83 Stoichiometric calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190

1.83.1 Stoichiometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

1.84 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

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1.85 End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

The Hydrosphere 197

1.86 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

1.87 Interactions of the hydrosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197

1.88 Exploring the Hydrosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

1.88.1 Investigation : Investigating the hydrosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

1.89 The Importance of the Hydrosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

1.90 Threats to the Hydrosphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

1.90.1 Discussion : Creative water conservation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200

1.90.2 Investigation: Building of dams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

1.90.3 Group Project : School Action Project . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

1.91 How pure is our water? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

1.91.1 Experiment: Water purity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

1.91.2 Project: water purification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

1.92 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

1.93 End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

II Physics 205

Vectors 207

1.94 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

1.95 Scalars and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

1.96 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

1.96.1 Mathematical Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

1.96.2 Graphical Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

1.97 Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

1.97.1 Relative Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

1.97.2 Compass Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

1.97.3 Bearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

Scalars and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

1.98 Drawing Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

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1.98.1 Drawing Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212

1.99 Mathematical Properties of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

1.99.1 Adding Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

1.99.2 Subtracting Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216

1.99.3 Scalar Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

1.100Techniques of Vector Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

1.100.1 Graphical Techniques . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

The Head-to-Tail Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218

The Parallelogram Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

1.100.2 Algebraic Addition and Subtraction of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

Vectors in a Straight Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

Resultant Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

A More General Algebraic technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

More Resultant Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

1.101Components of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230

1.101.1 Block on an incline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232

1.101.2 Worked example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

1.101.3 Vector addition using components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

Adding and Subtracting Components of Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

Vector Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

1.101.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

1.101.5 End of chapter exercises: Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240

1.101.6 End of chapter exercises: Vectors - Long questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242

Mechanical Energy 245

1.102Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

1.103Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245

1.103.1 Gravitational Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

1.104Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248

1.104.1 Checking units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250

Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

1.105Mechanical Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

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1.105.1 Conservation of Mechanical Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

1.105.2 Using the Law of Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253

Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

1.106Energy graphs - (Not in CAPS, included for completeness) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256

1.107Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

1.108End of Chapter Exercises: Mechanical Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

Motion in One Dimension 261

1.109Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

1.110Reference Point, Frame of Reference and Position . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

1.110.1 Frames of Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262

1.110.2 Position . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

Discussion : Reference Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

Position . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

1.111Displacement and Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265

1.111.1 Interpreting Direction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

1.111.2 Differences between Distance and Displacement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

Point of Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268

1.112Speed, Average Velocity and Instantaneous Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268

1.112.1 Differences between Speed and Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274

Displacement and related quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274

Investigation : An Exercise in Safety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276

1.113Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277

1.113.1 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

1.114Description of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

1.114.1 Stationary Object . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280

1.114.2 Motion at Constant Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281

Velocity and acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

Experiment : Motion at constant velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

1.114.3 Motion at Constant Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285

Velocity from Acceleration vs. Time Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

1.114.4 Summary of Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287

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1.114.5 Experiment: Position versus time using a ticker timer . . . . . . . . . . . . . . . . . . . . . . . . . . . 288

1.114.6 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289

1.114.7 Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295

1.115Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296

1.115.1 Finding the Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297

Derivation of (15.50) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297

Derivation of (15.51) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298

Derivation of (15.52) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

Derivation of (15.53) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

1.116Applications in the Real-World . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302

1.117Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303

1.118End of Chapter Exercises: Motion in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304

Transverse Pulses 311

1.119Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311

1.120What is a medium? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311

1.121What is a pulse? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312

1.121.1 Investigation : Observation of Pulses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312

1.121.2 Pulse Length and Amplitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312

Investigation : Pulse Length and Amplitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313

1.121.3 Pulse Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314

Pulse Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315

1.122Superposition of Pulses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315

1.122.1 Experiment: Constructive and destructive interference . . . . . . . . . . . . . . . . . . . . . . . . . . . 318

1.122.2 Problems Involving Superposition of Pulses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319

1.123Graphs of Position and Velocity (Not in CAPS - Included for Completeness) . . . . . . . . . . . . . . . . . . . 321

1.123.1 Motion of a Particle of the Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

Investigation : Drawing a position-time graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322

Investigation : Drawing a velocity-time graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323

1.123.2 Motion of the Pulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324

Travelling Pulse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326

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1.124Transmission and Reflection of a Pulse at a Boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327

1.124.1 Investigation : Two ropes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328

1.124.2 Pulses at a Boundary I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329

1.125Reflection of a Pulse from Fixed and Free Ends (not in CAPS - included for completeness) . . . . . . . . . . . 330

1.125.1 Reflection of a Pulse from a Fixed End . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330

Investigation : Reflection of a Pulse from a Fixed End . . . . . . . . . . . . . . . . . . . . . . . . . . . 330

1.125.2 Reflection of a Pulse from a Free End . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331

Investigation : Reflection of a Pulse from a Free End . . . . . . . . . . . . . . . . . . . . . . . . . . . 331

Pulses at a Boundary II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331

1.126Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332

1.127Exercises - Transverse Pulses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332

Transverse Waves 335

1.128Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335

1.129What is a transverse wave? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335

1.129.1 Investigation : Transverse Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

1.129.2 Peaks and Troughs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

1.129.3 Amplitude and Wavelength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

Investigation : Amplitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338

Investigation : Wavelength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339

1.129.4 Points in Phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341

Investigation : Points in Phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341

1.129.5 Period and Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342

1.129.6 Speed of a Transverse Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343

Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345

1.130Graphs of Particle Motion (Not in CAPS - Included for Interest) . . . . . . . . . . . . . . . . . . . . . . . . . . 347

1.130.1 Mathematical Description of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350

1.130.2 Graphs of Particle Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351

1.131Standing Waves and Boundary Conditions (Not in CAPS - Included for Interest) . . . . . . . . . . . . . . . . . 351

1.131.1 Reflection of a Transverse Wave from a Fixed End . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351

1.131.2 Reflection of a Transverse Wave from a Free End . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352

1.131.3 Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352

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Investigation : Creating Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352

1.131.4 Nodes and Anti-nodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356

1.131.5 Wavelengths of Standing Waves with Fixed and Free Ends . . . . . . . . . . . . . . . . . . . . . . . . 357

One Node . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358

Two Nodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358

Three Nodes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359

1.131.6 Superposition and Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360

Superposition and Interference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361

1.132Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363

1.133Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363

Longitudinal Waves 365

1.134Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365

1.135What is a longitudinal wave? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365

1.135.1 Investigation : Investigating longitudinal waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 366

1.136Characteristics of Longitudinal Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367

1.136.1 Compression and Rarefaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367

1.136.2 Wavelength and Amplitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368

1.136.3 Period and Frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369

1.136.4 Speed of a Longitudinal Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369

1.137Sound Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 371

1.138Seismic Waves - (Not Included in CAPS - Included for Interest) . . . . . . . . . . . . . . . . . . . . . . . . . . 371

1.139 Graphs of Particle Position, Displacement, Velocity and Acceleration (Not Included in CAPS - Included for

Completeness) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372

1.140Summary - Longitudinal Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374

1.141Exercises - Longitudinal Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374

Sound 377

1.142Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377

1.143Characteristics of a Sound Wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377

1.143.1 Pitch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378

Investigation : Range of Wavelengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

1.143.2 Loudness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

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1.143.3 Tone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

1.144Speed of Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

1.144.1 Sound frequency and amplitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380

1.145Physics of the Ear and Hearing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380

1.146Ultrasound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381

1.147SONAR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382

1.147.1 Echolocation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383

1.148Intensity of Sound (Not Included in CAPS - Advanced) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384

1.148.1 Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384

1.148.2 dB Scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384

1.148.3 Discussion : Importance of Safety Equipment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385

1.149Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385

1.150Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386

Electromagnetic Radiation 391

1.151Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391

1.152Particle/Wave Nature of Electromagnetic Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391

1.152.1 Particle/wave nature of electromagnetic radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392

1.153The wave nature of electromagnetic radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392

1.154Electromagnetic spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393

1.154.1 Dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393

1.154.2 Wave Nature of EM Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394

1.154.3 EM Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396

1.155The particle nature of electromagnetic radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396

1.155.1 Exercise - particle nature of EM waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397

1.156Penetrating ability of electromagnetic radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398

1.156.1 Ultraviolet(UV) radiation and the skin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398

What makes a good sunscreen? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398

1.156.2 Ultraviolet radiation and the eyes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399

1.156.3 X-rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399

1.156.4 Gamma-rays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399

Cellphones and electromagnetic radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399

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1.156.5 Exercise - Penetrating ability of EM radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400

1.157Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400

1.158End of chapter exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400

Magnetism 403

1.159Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

1.160Magnetic fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403

1.160.1 Investigation : Ferromagnetic materials and magnetisation . . . . . . . . . . . . . . . . . . . . . . . . 404

1.161Permanent magnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405

1.161.1 The poles of permanent magnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405

1.161.2 Magnetic attraction and repulsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405

1.161.3 Representing magnetic fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407

Investigation : Field around a Bar Magnet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407

Investigation : Field around a Pair of Bar Magnets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408

Ferromagnetism and Retentivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409

1.162The compass and the earth’s magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409

1.162.1 The earth’s magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410

Phenomena related to the Earth’s magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411

The importance of the magnetic field to life on Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . 411

Aurorae (pronounced Or-roar-ee) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412

1.163Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412

1.164End of chapter exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412

Electrostatics 415

1.165Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

1.166Two kinds of charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

1.167Unit of charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415

1.168Conservation of charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 416

1.169Force between Charges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417

1.169.1 Experiment : Electrostatic Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417

1.170Conductors and insulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420

1.170.1 The electroscope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422

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Grounding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423

1.171Attraction between charged and uncharged objects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424

1.171.1 Polarisation of Insulators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424

1.172Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425

1.173End of chapter exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425

Electric Circuits 429

1.174Electric Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429

1.174.1 Discussion : Uses of electricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429

1.174.2 Closed circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430

Experiment : Closed circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430

1.174.3 Representing electric circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430

Components of electrical circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 430

Circuit diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432

Series and parallel circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433

Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435

Discussion : Alternative Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 436

1.175Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437

1.175.1 Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437

1.175.2 Potential difference and emf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438

EMF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 438

Experiment : Investigate the emf and potential difference of a battery . . . . . . . . . . . . . . . . . . 439

1.176Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444

1.176.1 Flow of Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444

1.176.2 Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444

Exercises: Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445

1.176.3 Measuring voltage and current in circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447

Voltmeter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447

Ammeter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448

Impact of meters on circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449

Investigation : Using meters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449

1.177Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449

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1.177.1 What causes resistance? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450

Why do batteries go flat? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450

1.177.2 Resistors in electric circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450

Resistors in series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 450

Equivalent Series Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452

Experiment : Current in Series Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453

Resistors in parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455

Equivalent parallel resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456

Experiment : Current in Parallel Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 457

Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459

1.178 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459

1.179Exercises - Electric circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460

Safety 463

.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

.2 General safety rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

.3 Hazard signs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463

.4 Notes and information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464

Attributions 477

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Chapter 1

Units

1.1 Introduction

(section shortcode: U10000 )

Imagine you had to make curtains and needed to buy fabric. The shop assistant would need to know how much fabric you

needed. Telling her you need fabric 2 wide and 6 long would be insufficient — you have to specify the unit (i.e. 2 metres wide

and 6 metres long). Without the unit the information is incomplete and the shop assistant would have to guess. If you were

making curtains for a doll’s house the dimensions might be 2 centimetres wide and 6 centimetres long!

It is not just lengths that have units, all physical quantities have units (e.g. time, temperature, distance, etc.).

Definition: Physical Quantity

A physical quantity is anything that you can measure. For example, length, temperature, distance

and time are physical quantities.

1.2 Unit Systems

(section shortcode: U10001 )

1.2.1 SI Units

We will be using the SI units in this course. SI units are the internationally agreed upon units. Historically these units are

based on the metric system which was developed in France at the time of the French Revolution.

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Definition: SI Units

The name SI units comes from the French Système International d’Unités, which means international

system of units.

There are seven base SI units. These are listed in Table 1. All physical quantities have units which can be built from these

seven base units. So, it is possible to create a different set of units by defining a different set of base units.

These seven units are called base units because none of them can be expressed as combinations of the other six. This is

identical to bricks and concrete being the base units of a building. You can build different things using different combinations

of bricks and concrete. The 26 letters of the alphabet are the base units for a language like English. Many different words can

be formed by using these letters.

Base quantity Name Symbol

length metre m

mass kilogram kg

time second s

electric current ampere A

temperature kelvin K

amount of substance mole mol

luminous intensity candela cd

Table 1: SI Base Units

1.2.2 The Other Systems of Units

The SI Units are not the only units available, but they are most widely used. In Science there are three other sets of units that

can also be used. These are mentioned here for interest only.

1.2.3 c.g.s. Units

In the c.g.s. system, the metre is replaced by the centimetre and the kilogram is replaced by the gram. This is a simple

change but it means that all units derived from these two are changed. For example, the units of force and work are different.

These units are used most often in astrophysics and atomic physics.

1.2.4 Imperial Units

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CHAPTER 1. UNITS 1.3

Imperial units arose when kings and queens decided the measures that were to be used in the land. All the imperial base

units, except for the measure of time, are different to those of SI units. This is the unit system you are most likely to encounter

if SI units are not used. Examples of imperial units are pounds, miles, gallons and yards. These units are used by the

Americans and British. As you can imagine, having different units in use from place to place makes scientific communication

very difficult. This was the motivation for adopting a set of internationally agreed upon units.

1.2.5 Natural Units

This is the most sophisticated choice of units. Here the most fundamental discovered quantities (such as the speed of light)

are set equal to 1. The argument for this choice is that all other quantities should be built from these fundamental units. This

system of units is used in high energy physics and quantum mechanics.

1.3 Writing Units as Words or Symbols

(section shortcode: U10002 )

Unit names are always written with a lowercase first letter, for example, we write metre and litre. The symbols or abbreviations

of units are also written with lowercase initials, for example m for metre and ℓ for litre. The exception to this rule is if the unit is

named after a person, then the symbol is a capital letter. For example, the kelvin was named after Lord Kelvin and its symbol

is K. If the abbreviation of the unit that is named after a person has two letters, the second letter is lowercase, for example Hz

for hertz.

1.3.1 Naming of Units

For the following symbols of units that you will come across later in this book, write whether you think the unit is named after

a person or not.

1. J (joule)

2. ℓ (litre)

3. N (newton)

4. mol (mole)

5. C (coulomb)

6. lm (lumen)

7. m (metre)

8. bar (bar)

Find the answers with the shortcodes:

(1.) lOX

1.4 Combinations of SI Base Units

(section shortcode: U10003 )

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1.5 CHAPTER 1. UNITS

To make working with units easier, some combinations of the base units are given special names, but it is always correct to

reduce everything to the base units. Table 2 lists some examples of combinations of SI base units that are assigned special

names. Do not be concerned if the formulae look unfamiliar at this stage - we will deal with each in detail in the chapters

ahead (as well as many others)!

It is very important that you are able to recognise the units correctly. For instance, the newton (N) is another name for the

kilogram metre per second squared (kg·m·s−2), while the kilogram metre squared per second squared (kg·m2·s−2) is

called the joule (J).

Quantity Formula Unit Expressed in Base Units Name of Combination

Force ma kg·m·s−2 N (newton)

Frequency 1

Ts−1 Hz (hertz)

Work Fs kg·m2·s−2 J (joule)

Table 2: Some examples of combinations of SI base units assigned special names

TIP: When writing combinations of base SI units, place a dot (·) between the units to indicate

that different base units are used. For example, the symbol for metres per second is correctly

written as m·s−1, and not as ms−1 or m/s. Although the last two options will be accepted in tests

and exams, we will only use the first one in this book.

1.5 Rounding, Scientific Notation and Significant Figures

(section shortcode: U10004 )

1.5.1 Rounding Off

Certain numbers may take an infinite amount of paper and ink to write out. Not only is that impossible, but writing numbers

out to a high precision (many decimal places) is very inconvenient and rarely gives better answers. For this reason we often

estimate the number to a certain number of decimal places. Rounding off or approximating a decimal number to a given

number of decimal places is the quickest way to approximate a number. For example, if you wanted to round-off 2, 6525272

to three decimal places then you would first count three places after the decimal. 2, 652|5272 All numbers to the right of | are

ignored after you determine whether the number in the third decimal place must be rounded up or rounded down. You round

up the final digit (make the digit one more) if the first digit after the | was greater or equal to 5 and round down (leave the digit

alone) otherwise. So, since the first digit after the | is a 5, we must round up the digit in the third decimal place to a 3 and the

final answer of 2, 6525272 rounded to three decimal places is 2,653.

Exercise 1: Rounding-off Round off π = 3, 141592654... to 4 decimal places.

Solution to Exercise

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CHAPTER 1. UNITS 1.5

Step 1. π = 3, 1415|92654...Step 2. The last digit of π = 3, 1415|92654... must be rounded up because there

is a 9 after the |.Step 3. π = 3, 1416 rounded to 4 decimal places.

Exercise 2: Rounding-off Round off 9, 191919... to 2 decimal places

Solution to Exercise

Step 1. 9, 19|1919...Step 2. The last digit of 9, 19|1919... must be rounded down because there

is a 1 after the |.Step 3. Answer = 9,19 rounded to 2 decimal places.

1.5.2 Error Margins

In a calculation that has many steps, it is best to leave the rounding off right until the end. For example, Jack and

Jill walk to school. They walk 0,9 kilometers to get to school and it takes them 17 minutes. We can calculate their

speed in the following two ways.

Method 1:

timeinhours = 17min60min

= 0, 283333333h(1)

speed = DistanceTime

= 0,9km0,28333333h

= 3, 176470588km · h−1

= 3, 18km · h−1

(2)

Method 2:

timeinhours = 17min60min

= 0, 28h(3)

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1.5 CHAPTER 1. UNITS

speed = DistanceTime

= 0,9km0,28h

= 3, 214285714km · h−1

= 3, 21km · h−1

(4)

You will see that we get two different answers. In Method 1 no rounding was done, but in Method 2, the time was

rounded to 2 decimal places. This made a big difference to the answer. The answer in Method 1 is more accurate

because rounded numbers were not used in the calculation. Always only round off your final answer.

1.5.3 Scientific Notation

In Science one often needs to work with very large or very small numbers. These can be written more easily in

scientific notation, in the general form

d× 10e (5)

where d is a decimal number between 0 and 10 that is rounded off to a few decimal places. e is known as the

exponent and is an integer. If e > 0 it represents how many times the decimal place in d should be moved to the

right. If e < 0, then it represents how many times the decimal place in d should be moved to the left. For example

3, 24 × 103 represents 3240 (the decimal moved three places to the right) and 3, 24 × 10−3 represents 0, 00324(the decimal moved three places to the left).

If a number must be converted into scientific notation, we need to work out how many times the number must be

multiplied or divided by 10 to make it into a number between 1 and 10 (i.e. the value of e) and what this number

between 1 and 10 is (the value of d). We do this by counting the number of decimal places the decimal comma

must move.

For example, write the speed of light in scientific notation, to two decimal places. The speed of light is 299 792

458 m·s−1. First, find where the decimal comma must go for two decimal places (to find d) and then count how

many places there are after the decimal comma to determine e.

In this example, the decimal comma must go after the first 2, but since the number after the 9 is 7, d = 3, 00.

e = 8 because there are 8 digits left after the decimal comma. So the speed of light in scientific notation, to two

decimal places is 3,00 × 108 m·s−1.

1.5.4 Significant Figures

In a number, each non-zero digit is a significant figure. Zeroes are only counted if they are between two non-zero

digits or are at the end of the decimal part. For example, the number 2000 has 1 significant figure (the 2), but

2000,0 has 5 significant figures. You estimate a number like this by removing significant figures from the number

(starting from the right) until you have the desired number of significant figures, rounding as you go. For example

6,827 has 4 significant figures, but if you wish to write it to 3 significant figures it would mean removing the 7 and

rounding up, so it would be 6,83.

6

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CHAPTER 1. UNITS 1.6. PREFIXES OF BASE UNITS

1.5.5 Using Significant Figures

1. Round the following numbers:

a. 123,517 ℓ to 2 decimal places

b. 14,328 km·h−1 to one decimal place

c. 0,00954 m to 3 decimal places

2. Write the following quantities in scientific notation:

a. 10130 Pa to 2 decimal places

b. 978,15 m·s−2 to one decimal place

c. 0,000001256 A to 3 decimal places

3. Count how many significant figures each of the quantities below has:

a. 2,590 km

b. 12,305 mℓc. 7800 kg

Find the answers with the shortcodes:

(1.) lOl (2.) lO5 (3.) lON

1.6 Prefixes of Base Units

(section shortcode: U10005 )

Now that you know how to write numbers in scientific notation, another important aspect of units is the prefixes

that are used with the units.

Definition: Prefix

A prefix is a group of letters that are placed in front of a word. The effect of the prefix is to

change meaning of the word. For example, the prefix un is often added to a word to mean

not, as in unnecessary which means not necessary.

In the case of units, the prefixes have a special use. The kilogram (kg) is a simple example. 1 kg is equal to 1

000 g or 1× 103 g. Grouping the 103 and the g together we can replace the 103 with the prefix k (kilo). Therefore

the k takes the place of the 103. The kilogram is unique in that it is the only SI base unit containing a prefix.

In Science, all the prefixes used with units are some power of 10. Table 3 lists some of these prefixes. You will

not use most of these prefixes, but those prefixes listed in bold should be learnt. The case of the prefix symbol

is very important. Where a letter features twice in the table, it is written in uppercase for exponents bigger than

one and in lowercase for exponents less than one. For example M means mega (106) and m means milli (10−3).

7

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1.6. PREFIXES OF BASE UNITS CHAPTER 1. UNITS

Prefix Symbol Exponent Prefix Symbol Exponent

yotta Y 1024 yocto y 10−24

zetta Z 1021 zepto z 10−21

exa E 1018 atto a 10−18

peta P 1015 femto f 10−15

tera T 1012 pico p 10−12

giga G 109 nano n 10−9

mega M 106 micro µ 10−6

kilo k 103 milli m 10−3

hecto h 102 centi c 10−2

deca da 101 deci d 10−1

Table 3: Unit Prefixes

TIP: There is no space and no dot between the prefix and the symbol for the unit.

Here are some examples of the use of prefixes:

• 40000 m can be written as 40 km (kilometre)

• 0,001 g is the same as 1× 10−3 g and can be written as 1 mg (milligram)

• 2, 5× 106 N can be written as 2,5 MN (meganewton)

• 250000 A can be written as 250 kA (kiloampere) or 0,250 MA (megaampere)

• 0,000000075 s can be written as 75 ns (nanoseconds)

• 3× 10−7 mol can be rewritten as 0, 3× 10−6 mol, which is the same as 0,3 µmol (micromol)

1.6.1 Using Scientific Notation

1. Write the following in scientific notation using Table 3 as a reference.

a. 0,511 MV

b. 10 cℓc. 0,5 µm

d. 250 nm

e. 0,00035 hg

2. Write the following using the prefixes in Table 3.

a. 1,602 ×10−19 C

b. 1,992 ×106 J

c. 5,98 ×104 N

d. 25 ×10−4 A

8

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CHAPTER 1. UNITS 1.7. THE IMPORTANCE OF UNITS

e. 0,0075 ×106 m

Find the answers with the shortcodes:

(1.) lOR (2.) lOn

1.7 The Importance of Units

(section shortcode: U10006 )

Without units much of our work as scientists would be meaningless. We need to express our thoughts clearly and

units give meaning to the numbers we measure and calculate. Depending on which units we use, the numbers

are different. For example if you have 12 water, it means nothing. You could have 12 ml of water, 12 litres of

water, or even 12 bottles of water. Units are an essential part of the language we use. Units must be specified

when expressing physical quantities. Imagine that you are baking a cake, but the units, like grams and millilitres,

for the flour, milk, sugar and baking powder are not specified!

1.7.1 Investigation : Importance of Units

Work in groups of 5 to discuss other possible situations where using the incorrect set of units can be to your

disadvantage or even dangerous. Look for examples at home, at school, at a hospital, when travelling and in a

shop.

9

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1.8. HOW TO CHANGE UNITS CHAPTER 1. UNITS

1.7.2 Case Study : The importance of units

Read the following extract from CNN News 30 September 1999 and answer the questions below.

NASA: Human error caused loss of Mars orbiter November 10, 1999

Failure to convert English measures to metric values caused the loss of the Mars Climate Orbiter, a spacecraft

that smashed into the planet instead of reaching a safe orbit, a NASA investigation concluded Wednesday. The

Mars Climate Orbiter, a key craft in the space agency’s exploration of the red planet, vanished on 23 September

after a 10 month journey. It is believed that the craft came dangerously close to the atmosphere of Mars, where it

presumably burned and broke into pieces. An investigation board concluded that NASA engineers failed to con-

vert English measures of rocket thrusts to newton, a metric system measuring rocket force. One English pound

of force equals 4,45 newtons. A small difference between the two values caused the spacecraft to approach Mars

at too low an altitude and the craft is thought to have smashed into the planet’s atmosphere and was destroyed.

The spacecraft was to be a key part of the exploration of the planet. From its station about the red planet, the

Mars Climate Orbiter was to relay signals from the Mars Polar Lander, which is scheduled to touch down on Mars

next month. “The root cause of the loss of the spacecraft was a failed translation of English units into metric units

and a segment of ground-based, navigation-related mission software,” said Arthus Stephenson, chairman of the

investigation board. Questions:

1. Why did the Mars Climate Orbiter crash? Answer in your own words.

2. How could this have been avoided?

3. Why was the Mars Orbiter sent to Mars?

4. Do you think space exploration is important? Explain your answer.

1.8 How to Change Units

(section shortcode: U10007 )

It is very important that you are aware that different systems of units exist. Furthermore, you must be able to

convert between units. Being able to change between units (for example, converting from millimetres to metres)

is a useful skill in Science.

The following conversion diagrams will help you change from one unit to another.

Figure 1: The distance conversion table

10

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CHAPTER 1. UNITS 1.8. HOW TO CHANGE UNITS

If you want to change millimetre to metre, you divide by 1000 (follow the arrow from mm to m); or if you want to

change kilometre to millimetre, you multiply by 1000×1000.

The same method can be used to change millilitre to litre or kilolitre. Use Figure 2 to change volumes:

Figure 2: The volume conversion table

Exercise 3: Conversion 1 Express 3 800 mm in metres.

Solution to Exercise

Step 1. Use Figure 1 . Millimetre is on the left and metre in the middle.

Step 2. You need to go from mm to m, so you are moving from left to right.

Step 3. 3 800 mm ÷ 1000 = 3,8 m

Exercise 4: Conversion 2 Convert 4,56 kg to g.

Solution to Exercise

Step 1. Use Figure 1. Kilogram is the same as kilometre and gram the

same as metre.

Step 2. You need to go from kg to g, so it is from right to left.

Step 3. 4,56 kg × 1000 = 4560 g

11

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1.9. A SANITY TEST CHAPTER 1. UNITS

1.8.1 Two other useful conversions

Very often in Science you need to convert speed and temperature. The following two rules will help you do this:

Converting speed When converting km·h−1 to m·s−1you divide by 3,6.

For example 72 km·h−1÷ 3,6 = 20 m·s−1.

When converting m·s−1to km·h−1, you multiply by 3,6. For example 30 m·s−1×3,6 = 108 km·h−1.

Converting temperature Converting between the kelvin and celsius temperature scales is easy. To convert from

celsius to kelvin add 273. To convert from kelvin to celsius subtract 273. Representing the kelvin temperature by

TK and the celsius temperature by ToC ,

TK = ToC + 273 (6)

1.9 A sanity test

(section shortcode: U10008 )

A sanity test is a method of checking whether an answer makes sense. All we have to do is to take a careful look

at our answer and ask the question Does the answer make sense?

Imagine you were calculating the number of people in a classroom. If the answer you got was 1 000 000 people

you would know it was wrong — it is not possible to have that many people in a classroom. That is all a sanity

test is — is your answer insane or not?

It is useful to have an idea of some numbers before we start. For example, let us consider masses. An average

person has a mass around 70 kg, while the heaviest person in medical history had a mass of 635 kg. If you

ever have to calculate a person’s mass and you get 7 000 kg, this should fail your sanity check — your answer is

insane and you must have made a mistake somewhere. In the same way an answer of 0.01 kg should fail your

sanity test.

The only problem with a sanity check is that you must know what typical values for things are. For example, finding

the number of learners in a classroom you need to know that there are usually 20–50 people in a classroom. If

you get and answer of 2500, you should realise that it is wrong.

1.9.1 The scale of the matter... :

Try to get an idea of the typical values for the following physical quantities and write your answers into the table:

12

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CHAPTER 1. UNITS 1.10. SUMMARY

Category Quantity Minimum Maximum

People mass

height

Transport speed of cars on freeways

speed of trains

speed of aeroplanes

distance between home and school

General thickness of a sheet of paper

height of a doorway

Table 4

1.10 Summary

(section shortcode: U10009 )

1. You need to know the seven base SI Units as listed in Table 1. Combinations of SI Units can have different

names.

2. Unit names and abbreviations are written with lowercase letter unless it is named after a person.

3. Rounding numbers and using scientific notation is important.

4. Table 3 summarises the prefixes used in Science.

5. Use figures Figure 1 and Figure 2 to convert between units.

1.11 End of Chapter Exercises

(section shortcode: U10010 )

1. Write down the SI unit for the each of the following quantities:

a. length

b. time

c. mass

d. quantity of matter

2. For each of the following units, write down the symbol and what power of 10 it represents:

a. millimetre

b. centimetre

c. metre

d. kilometre

13

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1.11. END OF CHAPTER EXERCISES CHAPTER 1. UNITS

3. For each of the following symbols, write out the unit in full and write what power of 10 it represents:

a. µg

b. mg

c. kg

d. Mg

4. Write each of the following in scientific notation, correct to 2 decimal places:

a. 0,00000123 N

b. 417 000 000 kg

c. 246800 A

d. 0,00088 mm

5. Rewrite each of the following, accurate to two decimal places, using the correct prefix where applicable:

a. 0,00000123 N

b. 417 000 000 kg

c. 246800 A

d. 0,00088 mm

6. For each of the following, write the measurement using the correct symbol for the prefix and the base unit:

a. 1,01 microseconds

b. 1 000 milligrams

c. 7,2 megameters

d. 11 nanolitre

7. The Concorde is a type of aeroplane that flies very fast. The top speed of the Concorde is 2 172 km·hr−1.

Convert the Concorde’s top speed to m·s−1.

8. The boiling point of water is 100 C. What is the boiling point of water in kelvin?

Find the answers with the shortcodes:

(1.) lOQ (2.) lOU (3.) lOP (4.) lOE (5.) lOm (6.) lOy

(7.) lOV (8.) lOp

14

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Part I

Chemistry

15

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Classification of Matter

1.12 Introduction

(section shortcode: C10000 )

All the objects that we see in the world around us, are made of matter. Matter makes up the air we breathe, the

ground we walk on, the food we eat and the animals and plants that live around us. Even our own human bodies

are made of matter!

Different objects can be made of different types of matter, or materials. For example, a cupboard (an object)

is made of wood, nails and hinges (the materials). The properties of the materials will affect the properties

of the object. In the example of the cupboard, the strength of the wood and metals make the cupboard strong

and durable. In the same way, the raincoats that you wear during bad weather, are made of a material that is

waterproof. The electrical wires in your home are made of metal because metals are a type of material that is

able to conduct electricity. It is very important to understand the properties of materials, so that we can use them

in our homes, in industry and in other applications. In this chapter, we will be looking at different types of materials

and their properties.

Some of the properties of matter that you should know are:

• Materials can be strong and resist bending (e.g. iron rods, cement) or weak (e.g. fabrics)

• Materials that conduct heat (e.g. metals) are called thermal conductors. Materials that conduct electricity

are electrical conductors.

• Brittle materials break easily. Materials that are malleable can be easily formed into different shapes.

Ductile materials are able to be formed into long wires.

• Magnetic materials have a magnetic field.

• Density is the mass per unit volume. An example of a dense material is concrete.

• The boiling and melting points of substance help us to classify substances as solids, liquids or gases at a

specific temperature.

The diagram below shows one way in which matter can be classified (grouped) according to its different proper-

ties. As you read further in this chapter, you will see that there are also other ways of classifying materials, for

example according to whether or not they are good electrical conductors.

17

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1.13. MIXTURES CHAPTER 1. CLASSIFICATION OF MATTER

Figure 2.1: The classification of matter

Discussion: Everyday materials: In groups of 3 or 4 look at the labels of medicines, food items, and any other

items that you use often. What can you tell about the material inside the container from the list of ingredients?

Why is it important to have a list of ingredients on the materials that we use? Do some research on the safety

data of the various compounds in the items that you looked at. Are the compounds in the items safe to use? In

the food items, what preservatives and additives are there? Are these preservatives and additives good for you?

Are there natural alternatives (natural alternatives are usually used by indigenous people groups)?

1.13 Mixtures

(section shortcode: C10001 )

We see mixtures all the time in our everyday lives. A stew, for example, is a mixture of different foods such as

meat and vegetables; sea water is a mixture of water, salt and other substances, and air is a mixture of gases

such as carbon dioxide, oxygen and nitrogen.

Definition: Mixture

A mixture is a combination of two or more substances, where these substances are not

bonded (or joined) to each other.

In a mixture, the substances that make up the mixture:

• are not in a fixed ratio Imagine, for example, that you have a 250 ml beaker of water. It doesn’t matter

whether you add 20 g, 40 g, 100 g or any other mass of sand to the water; it will still be called a mixture of

sand and water.

• keep their physical properties In the example we used of the sand and water, neither of these substances

has changed in any way when they are mixed together. Even though the sand is in water, it still has the

same properties as when it was out of the water.

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CHAPTER 1. CLASSIFICATION OF MATTER 1.13. MIXTURES

• can be separated by mechanical means To separate something by ’mechanical means’, means that there

is no chemical process involved. In our sand and water example, it is possible to separate the mixture by

simply pouring the water through a filter. Something physical is done to the mixture, rather than something

chemical.

Some other examples of mixtures include blood (a mixture of blood cells, platelets and plasma), steel (a mixture

of iron and other materials) and the gold that is used to make jewellery. The gold in jewellery is not pure gold but

is a mixture of metals. The amount of gold in the jewellery is measured in karats (24 karat would be pure gold,

while 18 karat is only 75% gold).

We can group mixtures further by dividing them into those that are heterogeneous and those that are homoge-

neous.

1.13.1 Heterogeneous mixtures

A heterogeneous mixture does not have a definite composition. Think of a pizza, that has a topping of cheese,

tomato, mushrooms and peppers (the topping is a mixture). Each slice will probably be slightly different from the

next because the toppings (the tomato, cheese, mushrooms and peppers) are not evenly distributed. Another

example would be granite, a type of rock. Granite is made up of lots of different mineral substances including

quartz and feldspar. But these minerals are not spread evenly through the rock and so some parts of the rock

may have more quartz than others. Another example is a mixture of oil and water. Although you may add one

substance to the other, they will stay separate in the mixture. We say that these heterogeneous mixtures are

non-uniform, in other words they are not exactly the same throughout.

Definition: Heterogeneous mixture

A heterogeneous mixture is one that is non-uniform and the different components of the

mixture can be seen.

1.13.2 Homogeneous mixtures

A homogeneous mixture has a definite composition, and specific properties. In a homogeneous mixture, the

different parts cannot be seen. A solution of salt dissolved in water is an example of a homogeneous mixture.

When the salt dissolves, it will spread evenly through the water so that all parts of the solution are the same, and

you can no longer see the salt as being separate from the water. Think also of a powdered drink that you mix with

water. Provided you give the container a good shake after you have added the powder to the water, the drink will

have the same sweet taste for anyone who drinks it, it won’t matter whether they take a sip from the top or from

the bottom. The air we breathe is another example of a homogeneous mixture since it is made up of different

gases which are in a constant ratio, and which can’t be distinguished from each other.

Definition: Homogeneous mixture

A homogeneous mixture is one that is uniform, and where the different components of the

mixture cannot be seen.

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1.14. PURE SUBSTANCES: ELEMENTS AND COMPOUNDS CHAPTER 1. CLASSIFICATION OF MATTER

An alloy is a homogeneous mixture of two or more elements, at least one of which is a metal, where the resulting

material has metallic properties. Alloys are usually made to improve the properties of the elements that make

them up. For example steel is much stronger than iron (which is the main component of steel).

Activity: Classifying materials: Look around your classroom or school. Make a list of all the different materials

that you see around you. Try to work out why a particular material was used. Can you classify all the different

materials used according to their properties? On your way to school or at home or in the shops, look at the

different materials that are used. Why are these materials chosen over other materials?

Activity: Making mixtures: Make mixtures of sand and water, potassium dichromate and water, iodine and

ethanol, iodine and water. Classify these as heterogeneous or homogeneous. Try to make mixtures using other

substances. Are the mixtures that you have made heterogeneous or homogeneous? Give reasons for your

choice.

1.13.3 Mixtures

1. Which of the following substances are mixtures?

a. tap water

b. brass (an alloy of copper and zinc)

c. concrete

d. aluminium

e. Coca cola

f. distilled water

2. In each of the examples above, say whether the mixture is homogeneous or heterogeneous.

Find the answers with the shortcodes:

(1.) llm

1.14 Pure Substances: Elements and Compounds

(section shortcode: C10002 )

Any material that is not a mixture, is called a pure substance. Pure substances include elements and com-

pounds. It is much more difficult to break down pure substances into their parts, and complex chemical methods

are needed to do this.

One way to determine if a substance is pure is to look at its melting or boiling point. Pure substances will have a

sharply defined melting or boiling point (i.e. the melting or boiling point will be a single temperature rather than a

range of temperatures.) Impure substances have a temperature range over which they melt or boil. We can also

use chromatography to determine if a substance is pure or not. Chromatography is the process of separating

substances into their individual components. If a substance is pure then chromatography will only produce one

substance at the end of the process. If a substance is impure then several substances will be seen at the end of

the process.

20

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CHAPTER 1. CLASSIFICATION OF MATTER 1.14. PURE SUBSTANCES: ELEMENTS AND COMPOUNDS

1.14.1 Activity: Smartie Chromatography

You will need filter paper (or chromatography paper), some smarties in different colours, water and an eye drop-

per.

Place a smartie in the center of a piece of filter paper. Carefully drop a few drops of water onto the smartie. You

should see rings of different colour forming around the smartie. Each colour is one of the individual colours that

are used to make up the colour of the smartie.

1.14.2 Elements

An element is a chemical substance that can’t be divided or changed into other chemical substances by any

ordinary chemical means. The smallest unit of an element is the atom.

Definition: Element

An element is a substance that cannot be broken down into other substances through

chemical means.

There are 112 officially named elements and about 118 known elements. Most of these are natural, but some are

man-made. The elements we know are represented in the Periodic Table of the Elements, where each element

is abbreviated to a chemical symbol. Examples of elements are magnesium (Mg), hydrogen (H), oxygen (O)

and carbon (C). On the Periodic Table you will notice that some of the abbreviations do not seem to match

the elements they represent. The element iron, for example, has the chemical formula Fe. This is because the

elements were originally given Latin names. Iron has the abbreviation Fe because its Latin name is ’ferrum’. In

the same way, sodium’s Latin name is ’natrium’ (Na) and gold’s is ’aurum’ (Au).

NOTE: Recently it was agreed that two more elements would be added to the list of

officially named elements. These are elements number 114 and 116. The proposed

name for element 114 is flerovium and for element 116 it is moscovium. This brings

the total number of officially named elements to 114.

1.14.3 Compounds

A compound is a chemical substance that forms when two or more elements combine in a fixed ratio. Water

(H2O), for example, is a compound that is made up of two hydrogen atoms for every one oxygen atom. Sodium

chloride (NaCl) is a compound made up of one sodium atom for every chlorine atom. An important characteristic

of a compound is that it has a chemical formula, which describes the ratio in which the atoms of each element

in the compound occur.

Definition: Compound

A substance made up of two or more elements that are joined together in a fixed ratio.

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1.14. PURE SUBSTANCES: ELEMENTS AND COMPOUNDS CHAPTER 1. CLASSIFICATION OF MATTER

Figure 2.2 might help you to understand the difference between the terms element, mixture and compound. Iron

(Fe) and sulphur (S) are two elements. When they are added together, they form a mixture of iron and sulphur.

The iron and sulphur are not joined together. However, if the mixture is heated, a new compound is formed,

which is called iron sulphide (FeS). In this compound, the iron and sulphur are joined to each other in a ratio of

1:1. In other words, one atom of iron is joined to one atom of sulphur in the compound iron sulphide.

Figure 2.2: Understanding the difference between a mixture and a compound

Figure 2.2 shows the microscopic representation of mixtures and compounds. In a microscopic representation

we use circles to represent different elements. To show a compound, we draw several circles joined together.

Mixtures are simply shown as two or more individual elements in the same box. The circles are not joined for a

mixture.

We can also use symbols to represent elements, mixtures and compounds. The symbols for the elements are

all found on the periodic table. Compounds are shown as two or more element names written right next to each

other. Subscripts may be used to show that there is more than one atom of a particular element. (e.g. H2O or

NaCl). Mixtures are written as: a mixture of element (or compound) A and element (or compound) B. (e.g. a

mixture of Fe and S).

One way to think of mixtures and compounds is to think of buildings. The building is a mixture of different building

materials (e.g. glass, bricks, cement, etc.). The building materials are all compounds. You can also think of the

elements as Lego blocks. Each Lego block can be added to other Lego blocks to make new structures, in the

same way that elements can combine to make compounds.

Activity: Using models to represent substances Use coloured balls and sticks to represent elements and

compounds. Think about the way that we represent substances microscopically. Would you use just one ball to

represent an element or many? Why?

Elements, mixtures and compounds

1. In the following table, tick whether each of the substances listed is a mixture or a pure substance. If it is a

mixture, also say whether it is a homogeneous or heterogeneous mixture.

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CHAPTER 1. CLASSIFICATION OF MATTER 1.15. GIVING NAMES AND FORMULAE TO SUBSTANCES

Substance Mixture or pure Homogeneous or heterogeneous mixture

fizzy colddrink

steel

oxygen

iron filings

smoke

limestone (CaCO3)

Table 2.1

2. In each of the following cases, say whether the substance is an element, a mixture or a compound.

a. Cub. iron and sulphur

c. Ald. H2SO4

e. SO3

Find the answers with the shortcodes:

(1.) lly (2.) llV

1.15 Giving names and formulae to substances

(section shortcode: C10003 )

Think about what you call your friends. Their full name is like the substances name and their nickname is like the

substances formulae. Without these names your friends would have no idea which of them you are referring to.

In the same way scientists like to have a consistent way of naming things and a short way of describing the thing

being named. This helps scientists to communicate efficiently.

It is easy to describe elements and mixtures. We simply use the names that we find on the periodic table for

elements and we use words to describe mixtures. But how are compounds named? In the example of iron

sulphide that was used earlier, which element is named first, and which ’ending’ is given to the compound name

(in this case, the ending is -ide)?

The following are some guidelines for naming compounds:

1. The compound name will always include the names of the elements that are part of it.

• A compound of iron (Fe) and sulphur (S) is iron sulphide (FeS)

• A compound of potassium (K) and bromine (Br) is potassium bromide (KBr)• A compound of sodium (Na) and chlorine (Cl) is sodium chlor ide (NaCl)

2. In a compound, the element that is on the left of the Periodic Table, is used first when naming the compound.

In the example of NaCl, sodium is a group 1 element on the left hand side of the table, while chlorine is in

group 7 on the right of the table. Sodium therefore comes first in the compound name. The same is true for

FeS and KBr.

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1.15. GIVING NAMES AND FORMULAE TO SUBSTANCES CHAPTER 1. CLASSIFICATION OF MATTER

3. The symbols of the elements can be used to represent compounds e.g. FeS, NaCl, KBr and H2O. These

are called chemical formulae. In the first three examples, the ratio of the elements in each compound is

1:1. So, for FeS, there is one atom of iron for every atom of sulphur in the compound. In the last example

(H2O) there are two atoms of hydrogen for every atom of oxygen in the compound.

4. A compound may contain compound ions. An ion is an atom that has lost (positive ion) or gained (negative

ion) electrons. Some of the more common compound ions and their formulae are given below.

Name of compound ion Formula

Carbonate CO2−3

Sulphate SO2−4

Hydroxide OH−

Ammonium NH+4

Nitrate NO−3

Hydrogen carbonate HCO−3

Phosphate PO3−4

Chlorate ClO−3

Cyanide CN−

Chromate CrO2−4

Permanganate MnO−4

Table 2.2

5. When there are only two elements in the compound, the compound is often given a suffix (ending) of -ide.

You would have seen this in some of the examples we have used so far. For compound ions, when a

non-metal is combined with oxygen to form a negative ion (anion) which then combines with a positive ion

(cation) from hydrogen or a metal, then the suffix of the name will be ...ate or ...ite. NO−3 for example, is a

negative ion, which may combine with a cation such as hydrogen (HNO3) or a metal like potassium (KNO3).

The NO−3 anion has the name nitrate. SO2−

3 in a formula is sulphite, e.g. sodium sulphite (Na2SO3).

SO2−4 is sulphate and PO3−

4 is phosphate.

6. Prefixes can be used to describe the ratio of the elements that are in the compound. You should know the

following prefixes: ’mono’ (one), ’di’ (two) and ’tri’ (three).

• CO (carbon monoxide) - There is one atom of oxygen for every one atom of carbon

• NO2 (nitrogen dioxide) - There are two atoms of oxygen for every one atom of nitrogen

• SO3 (sulphur trioxide) - There are three atoms of oxygen for every one atom of sulphur

The above guidelines also help us to work out the formula of a compound from the name of the compound.

TIP: When numbers are written as ’subscripts’ in compounds (i.e. they are written

below and to the right of the element symbol), this tells us how many atoms of that

element there are in relation to other elements in the compound. For example in ni-

trogen dioxide (NO2) there are two oxygen atoms for every one atom of nitrogen. In

sulphur trioxide (SO3), there are three oxygen atoms for every one atom of sulphur

in the compound. Later, when we start looking at chemical equations, you will notice

that sometimes there are numbers before the compound name. For example, 2H2Omeans that there are two molecules of water, and that in each molecule there are two

hydrogen atoms for every one oxygen atom.

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CHAPTER 1. CLASSIFICATION OF MATTER 1.15. GIVING NAMES AND FORMULAE TO SUBSTANCES

We can use these rules to help us name both ionic compounds and covalent compounds (more on these com-

pounds will be covered in a later chapter). However, covalent compounds are often given other names by scien-

tists to simplify the name. For example, if we have 2 hydrogen atoms and one oxygen atom the above naming

rules would tell us that the substance is dihydrogen monoxide. But this compound is better known as water! Or

if we had 1 carbon atom and 4 hydrogen atoms then the name would be carbon tetrahydride, but scientists call

this compound methane.

Exercise 2.1: Naming compounds What is the chemical name for

a. KMnO4

b. NH4Cl

Solution to Exercise

Step 1. For a) we have potassium and the permanganate ion. For b) we

have the ammonium ion and chlorine.

Step 2. For a) we list the potassium first and the permanganate ion second.

So a) is potassium permanganate. For b) we list the ammonium ion

first and change the ending of chlorine to -ide. So b) is ammonium

chloride.

Exercise 2.2 Write the chemical formulae for:

a. sodium sulphate

b. potassium chromate

Solution to Exercise

Step 1. In part a) we have Na+ (sodium) and SO2−4 (sulphate). In part b)

we have K+ (potassium) and CrO2−4 (chromate)

Step 2. In part a) the charge on sodium is +1 and the charge on sulphate

is −2, so we must have two sodiums for every sulphate. In part b)

the charge on potassium is +1 and the charge on chromate is −2,

so we must have two potassiums for every chromate.

Step 3. a) is Na2SO4 and b) is K2CrO4

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1.15. GIVING NAMES AND FORMULAE TO SUBSTANCES CHAPTER 1. CLASSIFICATION OF MATTER

1.15.1 Naming compounds

1. The formula for calcium carbonate is CaCO3.

a. Is calcium carbonate a mixture or a compound? Give a reason for your answer.

b. What is the ratio of Ca : C : O atoms in the formula?

2. Give the name of each of the following substances.

a. KBrb. HClc. KMnO4

d. NO2

e. NH4OHf. Na2SO4

3. Give the chemical formula for each of the following compounds.

a. potassium nitrate

b. sodium iodide

c. barium sulphate

d. nitrogen dioxide

e. sodium monosulphate

4. Refer to the diagram below, showing sodium chloride and water, and then answer the questions that follow.

a. What is the chemical formula for water?

b. What is the chemical formula for sodium chloride?

c. Label the water and sodium chloride in the diagram.

d. Give a description of the picture. Focus on whether there are elements or compounds and if it is a

mixture or not.

5. What is the formula of this molecule?

a. C6H2Ob. C2H6Oc. 2C6HOd. 2CH6O

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CHAPTER 1. CLASSIFICATION OF MATTER 1.16. METALS, METALLOIDS AND NON-METALS

Find the answers with the shortcodes:

(1.) llp (2.) lld (3.) llv (4.) llL (5.) llf

1.16 Metals, Metalloids and Non-metals

(section shortcode: C10004 )

The elements in the Periodic Table can also be divided according to whether they are metals, metalloids or

non-metals. On the right hand side of the Periodic Table you can draw a ’zigzag’ line (This line starts with Boron

(B) and goes down to Polonium (Po). This line separates all the elements that are metals from those that are

non-metals. Metals are found on the left of the line, and non-metals are those on the right. Along the line you find

the metalloids. You should notice that there are more metals then non-metals. Metals, metalloids and non-metals

all have their own specific properties.

1.16.1 Metals

Examples of metals include copper (Cu), zinc (Zn), gold (Au) and silver (Ag). On the Periodic Table, the metals

are on the left of the zig-zag line. There are a large number of elements that are metals. The following are some

of the properties of metals:

• Thermal conductors Metals are good conductors of heat. This makes them useful in cooking utensils such

as pots and pans.

• Electrical conductors Metals are good conductors of electricity. Metals can be used in electrical conducting

wires.

• Shiny metallic lustre Metals have a characteristic shiny appearance and so are often used to make jewellery.

• Malleable This means that they can be bent into shape without breaking.

• Ductile Metals (such as copper) can be stretched into thin wires, which can then be used to conduct

electricity.

• Melting point Metals usually have a high melting point and can therefore be used to make cooking pots and

other equipment that needs to become very hot, without being damaged.

You can see how the properties of metals make them very useful in certain applications.

Group Work : Looking at metals

1. Collect a number of metal items from your home or school. Some examples are listed below:

• hammer

• wire

• cooking pots

• jewellery

• nails

• coins

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1.17. ELECTRICAL CONDUCTORS, SEMI-CONDUCTORS AND INSULATORSCHAPTER 1. CLASSIFICATION OF MATTER

2. In groups of 3-4, combine your collection of metal objects.

3. What is the function of each of these objects?

4. Discuss why you think metal was used to make each object. You should consider the properties of metals

when you answer this question.

1.16.2 Non-metals

In contrast to metals, non-metals are poor thermal conductors, good electrical insulators (meaning that they do

not conduct electrical charge) and are neither malleable nor ductile. The non-metals are found on the right hand

side of the Periodic Table, and include elements such as sulphur (S), phosphorus (P), nitrogen (N) and oxygen

(O).

1.16.3 Metalloids

Metalloids or semi-metals have mostly non-metallic properties. One of their distinguishing characteristics is that

their conductivity increases as their temperature increases. This is the opposite of what happens in metals.

The metalloids include elements such as silicon (Si) and germanium (Ge). Notice where these elements are

positioned in the Periodic Table.

You should now be able to take any material and determine whether it is a metal, non-metal or metalloid simply

by using its properties.

1.17 Electrical conductors, semi-conductors and insulators

(section shortcode: C10005 )

An electrical conductor is a substance that allows an electrical current to pass through it. Electrical conductors

are usually metals. Copper is one of the best electrical conductors, and this is why it is used to make conducting

wire. In reality, silver actually has an even higher electrical conductivity than copper, but because silver is so

expensive, it is not practical to use it for electrical wiring because such large amounts are needed. In the overhead

power lines that we see above us, aluminium is used. The aluminium usually surrounds a steel core which adds

tensile strength to the metal so that it doesn’t break when it is stretched across distances. Occasionally gold is

used to make wire, not because it is a particularly good conductor, but because it is very resistant to surface

corrosion. Corrosion is when a material starts to deteriorate at the surface because of its reactions with the

surroundings, for example oxygen and water in the air.

An insulator is a non-conducting material that does not carry any charge. Examples of insulators would be plastic

and wood. Do you understand now why electrical wires are normally covered with plastic insulation? Semi-

conductors behave like insulators when they are cold, and like conductors when they are hot. The elements

silicon and germanium are examples of semi-conductors.

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CHAPTER 1. CLASSIFICATION OF MATTER1.17. ELECTRICAL CONDUCTORS, SEMI-CONDUCTORS AND INSULATORS

Definition: Conductors and insulators

A conductor allows the easy movement or flow of something such as heat or electrical

charge through it. Insulators are the opposite to conductors because they inhibit or reduce

the flow of heat, electrical charge, sound etc through them.

Think about the materials around you. Are they electrical conductors or not? Why are different materials used?

Think about the use of semiconductors in electronics? Can you think of why they are used there?

1.17.1 Experiment : Electrical conductivity

Aim:

To investigate the electrical conductivity of a number of substances

Apparatus:

• two or three cells

• light bulb

• crocodile clips

• wire leads

• a selection of test substances (e.g. a piece of plastic, aluminium can, metal pencil sharpener, magnet,

wood, chalk).

Method:

1. Set up the circuit as shown above, so that the test substance is held between the two crocodile clips. The

wire leads should be connected to the cells and the light bulb should also be connected into the circuit.

2. Place the test substances one by one between the crocodile clips and see what happens to the light bulb.

Results:

Record your results in the table below:

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1.18. THERMAL CONDUCTORS AND INSULATORS CHAPTER 1. CLASSIFICATION OF MATTER

Test substance Metal/non-metal Does the light bulb glow? Conductor or insulator

Table 2.3

Conclusions:

In the substances that were tested, the metals were able to conduct electricity and the non-metals were not.

Metals are good electrical conductors and non-metals are not.

The following simulation allows you to work through the above activity. For this simulation use the grab bag option

to get materials to test. Set up the circuit as described in the activity.

(Simulation: lbK)

1.18 Thermal Conductors and Insulators

(section shortcode: C10006 )

A thermal conductor is a material that allows energy in the form of heat, to be transferred within the mate-

rial, without any movement of the material itself. An easy way to understand this concept is through a simple

demonstration.

1.18.1 Demonstration : Thermal conductivity

Aim:

To demonstrate the ability of different substances to conduct heat.

Apparatus:

You will need two cups (made from the same material e.g. plastic); a metal spoon and a plastic spoon.

Method:

• Pour boiling water into the two cups so that they are about half full.

• At the same time, place a metal spoon into one cup and a plastic spoon in the other.

• Note which spoon heats up more quickly

Results:

The metal spoon heats up faster than the plastic spoon. In other words, the metal conducts heat well, but the

plastic does not.

Conclusion: Metal is a good thermal conductor, while plastic is a poor thermal conductor. This explains why

cooking pots are metal, but their handles are often plastic or wooden. The pot itself must be metal so that heat

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CHAPTER 1. CLASSIFICATION OF MATTER 1.18. THERMAL CONDUCTORS AND INSULATORS

from the cooking surface can heat up the pot to cook the food inside it, but the handle is made from a poor thermal

conductor so that the heat does not burn the hand of the person who is cooking.

An insulator is a material that does not allow a transfer of electricity or energy. Materials that are poor thermal

conductors can also be described as being good thermal insulators.

NOTE: Water is a better thermal conductor than air and conducts heat away from

the body about 20 times more efficiently than air. A person who is not wearing a

wetsuit, will lose heat very quickly to the water around them and can be vulnerable

to hypothermia (this is when the body temperature drops very low). Wetsuits help to

preserve body heat by trapping a layer of water against the skin. This water is then

warmed by body heat and acts as an insulator. Wetsuits are made out of closed-

cell, foam neoprene. Neoprene is a synthetic rubber that contains small bubbles of

nitrogen gas when made for use as wetsuit material. Nitrogen gas has very low thermal

conductivity, so it does not allow heat from the body (or the water trapped between

the body and the wetsuit) to be lost to the water outside of the wetsuit. In this way a

person in a wetsuit is able to keep their body temperature much higher than they would

otherwise.

1.18.2 Investigation : A closer look at thermal conductivity

Look at the table below, which shows the thermal conductivity of a number of different materials, and then answer

the questions that follow. The higher the number in the second column, the better the material is at conducting

heat (i.e. it is a good thermal conductor). Remember that a material that conducts heat efficiently, will also lose

heat more quickly than an insulating material.

Material Thermal Conductivity (W ·m−1 ·K−1 )

Silver 429

Stainless steel 16

Standard glass 1.05

Concrete 0.9 - 2

Red brick 0.69

Water 0.58

Snow 0.25 - 0.5

Wood 0.04 - 0.12

Polystyrene 0.03

Air 0.024

Table 2.4

Use this information to answer the following questions:

1. Name two materials that are good thermal conductors.

2. Name two materials that are good insulators.

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1.19. MAGNETIC AND NON-MAGNETIC MATERIALS CHAPTER 1. CLASSIFICATION OF MATTER

3. Explain why:

a. cooler boxes are often made of polystyrene

b. homes that are made from wood need less internal heating during the winter months.

c. igloos (homes made from snow) are so good at maintaining warm temperatures, even in freezing

conditions.

NOTE: It is a known fact that well-insulated buildings need less energy for heating

than do buildings that have no insulation. Two building materials that are being used

more and more worldwide, are mineral wool and polystyrene. Mineral wool is a

good insulator because it holds air still in the matrix of the wool so that heat is not lost.

Since air is a poor conductor and a good insulator, this helps to keep energy within the

building. Polystyrene is also a good insulator and is able to keep cool things cool and

hot things hot. It has the added advantage of being resistant to moisture, mould and

mildew.

Remember that concepts such as conductivity and insulation are not only relevant in the building, industrial and

home environments. Think for example of the layer of blubber or fat that is found in some animals. In very cold

environments, fat and blubber not only provide protection, but also act as an insulator to help the animal keep its

body temperature at the right level. This is known as thermoregulation.

1.19 Magnetic and Non-magnetic Materials

(section shortcode: C10007 )

We have now looked at a number of ways in which matter can be grouped, such as into metals, semi-metals and

non-metals; electrical conductors and insulators, and thermal conductors and insulators. One way in which we

can further group metals, is to divide them into those that are magnetic and those that are non-magnetic.

Definition: Magnetism

Magnetism is one of the phenomena by which materials exert attractive or repulsive forces

on other materials.

A metal is said to be ferromagnetic if it can be magnetised (i.e. made into a magnet). If you hold a magnet

very close to a metal object, it may happen that its own electrical field will be induced and the object becomes

magnetic. Some metals keep their magnetism for longer than others. Look at iron and steel for example. Iron

loses its magnetism quite quickly if it is taken away from the magnet. Steel on the other hand will stay magnetic

for a longer time. Steel is often used to make permanent magnets that can be used for a variety of purposes.

Magnets are used to sort the metals in a scrap yard, in compasses to find direction, in the magnetic strips of

video tapes and ATM cards where information must be stored, in computers and TV’s, as well as in generators

and electric motors.

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CHAPTER 1. CLASSIFICATION OF MATTER1.20. SEPERATING MIXTURES - NOT IN CAPS - INCLUDED FOR COMPLETENESS

1.19.1 Investigation : Magnetism

You can test whether an object is magnetic or not by holding another magnet close to it. If the object is attracted

to the magnet, then it too is magnetic.

Find some objects in your classroom or your home and test whether they are magnetic or not. Then complete

the table below:

Object Magnetic or non-magnetic

Table 2.5

1.19.2 Group Discussion : Properties of materials

In groups of 4-5, discuss how our knowledge of the properties of materials has allowed society to:

• develop advanced computer technology

• provide homes with electricity

• find ways to conserve energy

1.20 Seperating mixtures - Not in CAPS - included for completeness

(section shortcode: C10008 )

Sometimes it is important to be able to separate a mixture. There are lots of different ways to do this. These are

some examples:

• Filtration A piece of filter paper in a funnel can be used to separate a mixture of sand and water.

• Heating / evaporation Heating a solution causes the liquid (normally water) to evaporate, leaving the other

(solid) part of the mixture behind. You can try this using a salt solution.

• Centrifugation This is a laboratory process which uses the centrifugal force of spinning objects to separate

out the heavier substances from a mixture. This process is used to separate the cells and plasma in blood.

When the test tubes that hold the blood are spun round in the machine, the heavier cells sink to the bottom

of the test tube. Can you think of a reason why it might be important to have a way of separating blood in

this way?

• Dialysis This is an interesting way of separating a mixture because it can be used in some important

applications. Dialysis works using a process called diffusion. Diffusion takes place when one substance

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1.20. SEPERATING MIXTURES - NOT IN CAPS - INCLUDED FOR COMPLETENESSCHAPTER 1. CLASSIFICATION OF MATTER

in a mixture moves from an area where it has a high concentration to an area where its concentration is

lower. When this movement takes place across a semi-permeable membrane it is called osmosis. A semi-

permeable membrane is a barrier that lets some things move across it, but not others. This process is very

important for people whose kidneys are not functioning properly, an illness called renal failure.

NOTE: Normally, healthy kidneys remove waste products from the blood. When a

person has renal failure, their kidneys cannot do this any more, and this can be life-

threatening. Using dialysis, the blood of the patient flows on one side of a semi-

permeable membrane. On the other side there will be a fluid that has no waste prod-

ucts but lots of other important substances such as potassium ions (K+) that the per-

son will need. Waste products from the blood diffuse from where their concentration is

high (i.e. in the person’s blood) into the ’clean’ fluid on the other side of the membrane.

The potassium ions will move in the opposite direction from the fluid into the blood.

Through this process, waste products are taken out of the blood so that the person

stays healthy.

1.20.1 Investigation : The separation of a salt solution

Aim:

To demonstrate that a homogeneous salt solution can be separated using physical methods.

Apparatus:

glass beaker, salt, water, retort stand, bunsen burner.

Method:

1. Pour a small amount of water (about 20 ml) into a beaker.

2. Measure a teaspoon of salt and pour this into the water.

3. Stir until the salt dissolves completely. This is now called a salt solution. This salt solution is a homogeneous

mixture.

4. Place the beaker on a retort stand over a bunsen burner and heat gently. You should increase the heat until

the water almost boils.

5. Watch the beaker until all the water has evaporated. What do you see in the beaker?

Results:

The water evaporates from the beaker and tiny grains of salt remain at the bottom. (You may also observe grains

of salt on the walls of the beaker.)

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CHAPTER 1. CLASSIFICATION OF MATTER 1.21. SUMMARY

Conclusion:

The salt solution, which is a homogeneous mixture of salt and water, has been separated using heating and

evaporation.

1.20.2 Discussion : Separating mixtures

Work in groups of 3-4

Imagine that you have been given a container which holds a mixture of sand, iron filings (small pieces of iron

metal), salt and small stones of different sizes. Is this a homogeneous or a heterogeneous mixture? In your

group, discuss how you would go about separating this mixture into the four materials that it contains. The

following presentation provides a summary of the classification of matter.

(Presentation: P10009 )

1.21 Summary

(section shortcode: C10010 )

• All the objects and substances that we see in the world are made of matter.

• This matter can be classified according to whether it is a mixture or a pure substance.

• A mixture is a combination of one or more substances that are not chemically bonded to each other.

Examples of mixtures are air (a mixture of different gases) and blood (a mixture of cells, platelets and

plasma).

• The main characteristics of mixtures are that the substances that make them up are not in a fixed ratio,

they keep their individual properties and they can be separated from each other using mechanical means.

• A heterogeneous mixture is non-uniform and the different parts of the mixture can be seen. An example

would be a mixture of sand and water.

• A homogeneous mixture is uniform, and the different components of the mixture can’t be seen. An

example would be a salt solution. A salt solution is a mixture of salt and water. The salt dissolves in the

water, meaning that you can’t see the individual salt particles. They are interspersed between the water

molecules. Another example is a metal alloy such as steel.

• Mixtures can be separated using a number of methods such as filtration, heating, evaporation, centrifuga-

tion and dialysis.

• Pure substances can be further divided into elements and compounds.

• An element is a substance that can’t be broken down into simpler substances through chemical means.

• All the elements are recorded in the Periodic Table of the Elements. Each element has its own chemical

symbol. Examples are iron (Fe), sulphur (S), calcium (Ca), magnesium (Mg) and fluorine (F).

• A compound is a substance that is made up of two or more elements that are chemically bonded to each

other in a fixed ratio. Examples of compounds are sodium chloride (NaCl), iron sulphide (FeS), calcium

carbonate (CaCO3) and water (H2O).

• When naming compounds and writing their chemical formula, it is important to know the elements that

are in the compound, how many atoms of each of these elements will combine in the compound and where

the elements are in the Periodic Table. A number of rules can then be followed to name the compound.

• Another way of classifying matter is into metals (e.g. iron, gold, copper), semi-metals (e.g. silicon and

germanium) and non-metals (e.g. sulphur, phosphorus and nitrogen).

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1.21. SUMMARY CHAPTER 1. CLASSIFICATION OF MATTER

• Metals are good electrical and thermal conductors, they have a shiny lustre, they are malleable and ductile,

and they have a high melting point. These properties make metals very useful in electrical wires, cooking

utensils, jewellery and many other applications.

• A further way of classifying matter is into electrical conductors, semi-conductors and insulators.

• An electrical conductor allows an electrical current to pass through it. Most metals are good electrical

conductors.

• An electrical insulator is not able to carry an electrical current. Examples are plastic, wood, cotton material

and ceramic.

• Materials may also be classified as thermal conductors or thermal insulators depending on whether or

not they are able to conduct heat.

• Materials may also be either magnetic or non-magnetic.

1.21.1 Summary

1. For each of the following multiple choice questions, choose one correct answer from the list provided.

a. Which of the following can be classified as a mixture:

a. sugar

b. table salt

c. air

d. iron

b. An element can be defined as:

a. A substance that cannot be separated into two or more substances by ordinary chemical (or

physical) means

b. A substance with constant composition

c. A substance that contains two or more substances, in definite proportion by weight

d. A uniform substance

2. Classify each of the following substances as an element, a compound, a solution (homogeneous mixture),

or a heterogeneous mixture: salt, pure water, soil, salt water, pure air, carbon dioxide, gold and bronze.

3. Look at the table below. In the first column (A) is a list of substances. In the second column (B) is a

description of the group that each of these substances belongs in. Match up the substance in Column A

with the description in Column B.

Column A Column B

iron a compound containing 2 elements

H2S a heterogeneous mixture

sugar solution a metal alloy

sand and stones an element

steel a homogeneous mixture

Table 2.6

4. You are given a test tube that contains a mixture of iron filings and sulphur. You are asked to weigh the

amount of iron in the sample.

a. Suggest one method that you could use to separate the iron filings from the sulphur.

b. What property of metals allows you to do this?

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CHAPTER 1. CLASSIFICATION OF MATTER 1.21. SUMMARY

5. Given the following descriptions, write the chemical formula for each of the following substances:

a. silver metal

b. a compound that contains only potassium and bromine

c. a gas that contains the elements carbon and oxygen in a ratio of 1:2

6. Give the names of each of the following compounds:

a. NaBrb. BaSO4

c. SO2

7. For each of the following materials, say what properties of the material make it important in carrying out its

particular function.

a. tar on roads

b. iron burglar bars

c. plastic furniture

d. metal jewellery

e. clay for building

f. cotton clothing

Find the answers with the shortcodes:

(1.) ll6 (2.) llF (3.) llG (4.) ll7 (5.) llA (6.) llo

(7.) lls (8.) llH

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1.21. SUMMARY CHAPTER 1. CLASSIFICATION OF MATTER

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States of matter and the kinetic molecular

theory

1.22 Introduction

(section shortcode: C10011 )

In this chapter we will explore the states of matter and then look at the kinetic molecular theory. Matter exists in

three states: solid, liquid and gas. We will also examine how the kinetic theory of matter helps explain boiling and

melting points as well as other properties of matter.

NOTE: When a gas is heated above a certain temperature the electrons in the atoms

start to leave the atoms. The gas is said to be ionised. When a gas is ionised it is

known as a plasma. Plasmas share many of the properties of gases (they have no fixed

volume and fill the space they are in). This is a very high energy state and plasmas

often glow. Ionisation is the process of moving from a gas to a plasma and deionisation

is the reverse process. We will not consider plasmas further in this chapter.

1.23 States of matter

(section shortcode: C10012 )

All matter is made up of particles. We can see this when we look at diffusion. Diffusion is the movement of

particles from a high concentration to a low concentration. Diffusion can be seen as a spreading out of particles

resulting in an even distribution of the particles. You can see diffusion when you place a drop of food colouring in

water. The colour slowly spreads out through the water. If matter were not made of particles then we would only

see a clump of colour when we put the food colouring in water, as there would be nothing that could move about

and mix in with the water. The composition of matter will be looked at in What are the objects around us made

of? (Chapter 6).

Diffusion is a result of the constant thermal motion of particles. In Section 3.3 (The Kinetic Theory of Matter) we

will talk more about the thermal motion of particles.

In 1828 Robert Brown observed that pollen grains suspended in water moved about in a rapid, irregular motion.

This motion has since become known as Brownian motion. Brownian motion is essentially diffusion of many

particles.

Matter exists in one of three states, namely solid, liquid and gas. Matter can change between these states by

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1.23. STATES OF MATTER CHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY

either adding heat or removing heat. This is known as a change of state. As we heat an object (e.g. water) it

goes from a solid to a liquid to a gas. As we cool an object it goes from a gas to a liquid to a solid. The changes

of state that you should know are:

• Melting is the process of going from solid to liquid.

• Boiling (or evaporation) is the process of going from liquid to gas.

• Freezing is the process of going from liquid to solid.

• Condensation is the process of going from gas to liquid.

• Occasionally (e.g. for carbon dioxide) we can go directly from solid to gas in a process called sublimation.

A solid has a fixed shape and volume. A liquid takes on the shape of the container that it is in. A gas completely

fills the container that it is in. See Section 2.3 (The Kinetic Theory of Matter) for more on changes of state.

If we know the melting and boiling point of a substance then we can say what state (solid, liquid or gas) it will be

in at any temperature.

1.23.1 Experiment: States of matter

Aim To investigate the heating and cooling curve of water.

Apparatus beakers, ice, bunsen burner, thermometer, water.

Method

• Place some ice in a beaker

• Measure the temperature of the ice and record it.

• After 10 s measure the temperature again and record it. Repeat every 10 s, until at least 1 minute after the

ice has melted.

• Heat some water in a beaker until it boils. Measure and record the temperature of the water.

• Remove the water from the heat and measure the temperature every 10 s, until the beaker is cool to touch

WARNING: Be careful when handling the beaker of hot water. Do not touch the beaker

with your hands, you will burn yourself.

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CHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY1.24. THE KINETIC THEORY OF MATTER

Results Record your results in the following table:

Temperature of ice Time (s) Temperature of water Time (s)

Table 3.1: Table of results

Plot a graph of temperature against time for the ice melting and the boiling water cooling.

Discussion and conclusion Discuss your results with others in your class. What conclusions can you draw? You

should find that the temperature of the ice increases until the first drops of liquid appear and then the temperature

remains the same, until all the ice is melted. You should also find that when you cool water down from boiling,

the temperature remains constant for a while, then starts decreasing.

In the above experiment, you investigated the heating and cooling curves of water. We can draw heating and

cooling curves for any substance. A heating curve of a substance gives the changes in temperature as we move

from a solid to a liquid to a gas. A cooling curve gives the changes in temperature as we move from gas to liquid

to solid. An important observation is that as a substance melts or boils, the temperature remains constant until

the substance has changed state. This is because all the heat energy goes into breaking or forming the forces

between the molecules.

The above experiment is one way of demonstrating the changes of state of a substance. Ice melting or water

boiling should be very familiar to you.

1.24 The Kinetic Theory of Matter

(section shortcode: C10013 )

The kinetic theory of matter helps us to explain why matter exists in different phases (i.e. solid, liquid and gas),

and how matter can change from one phase to the next. The kinetic theory of matter also helps us to understand

other properties of matter. It is important to realise that what we will go on to describe is only a theory. It cannot

be proved beyond doubt, but the fact that it helps us to explain our observations of changes in phase, and other

properties of matter, suggests that it probably is more than just a theory.

Broadly, the Kinetic Theory of Matter says that:

• Matter is made up of particles that are constantly moving.

• All particles have energy, but the energy varies depending on whether the substance is a solid, liquid or

gas. Solid particles have the least amount of energy and gas particles have the greatest amount of energy.

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1.24. THE KINETIC THEORY OF MATTERCHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY

• The temperature of a substance is a measure of the average kinetic energy of the particles.

• A change in phase may occur when the energy of the particles is changed.

• There are spaces between the particles of matter.

• There are attractive forces between particles and these become stronger as the particles move closer

together. These attractive forces will either be intramolecular forces (if the particles are atoms) or inter-

molecular forces (if the particles are molecules). When the particles are extremely close, repulsive forces

start to act.

Table 3.2 summarises the characteristics of the particles that are in each phase of matter.

Property of matter Solid Liquid Gas

Particles Atoms or molecules Atoms or molecules Atoms or molecules

Energy and movement of

particles

Low energy - particles vi-

brate around a fixed point

Particles have less en-

ergy than in the gas

phase

Particles have high en-

ergy and are constantly

moving

Spaces between parti-

cles

Very little space between

particles. Particles are

tightly packed together

Smaller spaces than in

gases, but larger spaces

than in solids

Large spaces because of

high energy

Attractive forces between

particles

Very strong forces.

Solids have a fixed

volume.

Stronger forces than in

gas. Liquids can be

poured.

Weak forces because of

the large distance be-

tween particles

Changes in phase Solids become liquids if

their temperature is in-

creased. In some cases

a solid may become a

gas if the temperature is

increased.

A liquid becomes a gas

if its temperature is in-

creased. It becomes a

solid if its temperature

decreases.

In general a gas be-

comes a liquid when it

is cooled. (In a few

cases a gas becomes

a solid when cooled).

Particles have less en-

ergy and therefore move

closer together so that

the attractive forces be-

come stronger, and the

gas becomes a liquid (or

a solid.)

Table 3.2: Table summarising the general features of solids, liquids and gases.

The following presentation is a brief summary of the above. Try to fill in the blank spaces before clicking onto the

next slide.

(Presentation: P10014 )

Let’s look at an example that involves the three phases of water: ice (solid), water (liquid) and water vapour (gas).

Note that in the Figure 3.2 below the molecules in the solid phase are represented by single spheres, but they

would in reality look like the molecules in the liquid and gas phase. Sometimes we represent molecules as single

spheres in the solid phase to emphasise the small amount of space between them and to make the drawing

simpler.

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CHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY1.24. THE KINETIC THEORY OF MATTER

Figure 3.2: The three phases of matter

Taking water as an example we find that in the solid phase the water molecules have very little energy and can’t

move away from each other. The molecules are held closely together in a regular pattern called a lattice. If the

ice is heated, the energy of the molecules increases. This means that some of the water molecules are able to

overcome the intermolecular forces that are holding them together, and the molecules move further apart to form

liquid water. This is why liquid water is able to flow, because the molecules are more free to move than they were

in the solid lattice. If the molecules are heated further, the liquid water will become water vapour, which is a gas.

Gas particles have lots of energy and are far away from each other. That is why it is difficult to keep a gas in a

specific area! The attractive forces between the particles are very weak and they are only loosely held together.

Figure 3.3 shows the changes in phase that may occur in matter, and the names that describe these processes.

Figure 3.3: Changes in phase

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1.25. INTRAMOLECULAR AND INTERMOLECULAR FORCES (NOT IN CAPS - INCLUDED FOR

COMPLETENESS) CHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY

1.25 Intramolecular and intermolecular forces (Not in CAPS - Included

for Completeness)

(section shortcode: C10015 )

When atoms join to form molecules, they are held together by chemical bonds. The type of bond, and the

strength of the bond, depends on the atoms that are involved. These bonds are called intramolecular forces

because they are bonding forces inside a molecule (’intra’ means ’within’ or ’inside’). Sometimes we simply call

these intramolecular forces chemical bonds.

Definition: Intramolecular force

The force between the atoms of a molecule, which holds them together.

Examples of the types of chemical bonds that can exist between atoms inside a molecule are shown below.

These will be looked at in more detail in Chemical bonding (Chapter 5).

• Covalent bond Covalent bonds exist between non-metal atoms e.g. There are covalent bonds between the

carbon and oxygen atoms in a molecule of carbon dioxide.

• Ionic bond Ionic bonds occur between non-metal and metal atoms e.g. There are ionic bonds between the

sodium and chlorine atoms in a molecule of sodium chloride.

• Metallic bond Metallic bonds join metal atoms e.g. There are metallic bonds between copper atoms in a

piece of copper metal.

Intermolecular forces are those bonds that hold molecules together. A glass of water for example, contains

many molecules of water. These molecules are held together by intermolecular forces. The strength of the

intermolecular forces is important because they affect properties such as melting point and boiling point. For

example, the stronger the intermolecular forces, the higher the melting point and boiling point for that substance.

The strength of the intermolecular forces increases as the size of the molecule increases.

Definition: Intermolecular force

A force between molecules, which holds them together.

The following diagram may help you to understand the difference between intramolecular forces and intermolec-

ular forces.

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CHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY

1.25. INTRAMOLECULAR AND INTERMOLECULAR FORCES (NOT IN CAPS - INCLUDED FOR

COMPLETENESS)

Figure 3.4: Two representations showing the intermolecular and intramolecular forces in water: space-filling

model and structural formula.

It should be clearer now that there are two types of forces that hold matter together. In the case of water,

there are intramolecular forces that hold the two hydrogen atoms to the oxygen atom in each molecule of water

(these are the solid lines in the above diagram). There are also intermolecular forces between each of these

water molecules. These intermolecular forces join the hydrogen atom from one molecule to the oxygen atom of

another molecule (these are the dashed lines in the above figure). As mentioned earlier, these forces are very

important because they affect many of the properties of matter such as boiling point, melting point and a number

of other properties. Before we go on to look at some of these examples, it is important that we first take a look at

the Kinetic Theory of Matter.

TIP: To help you remember that intermolecular means between molecules, remember

that international means between nations.

1.25.1 Intramolecular and intermolecular forces

1. Using ammonia gas as an example...

a. Explain what is meant by an intramolecular force or chemical bond.

b. Explain what is meant by an intermolecular force.

2. Draw a diagram showing three molecules of carbon dioxide. On the diagram, show where the intramolecular

and intermolecular forces are.

3. Why is it important to understand the types of forces that exist between atoms and between molecules?

Try to use some practical examples in your answer.

Find the answers with the shortcodes:

(1.) lin (2.) liQ (3.) liU

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1.26. THE PROPERTIES OF MATTERCHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY

1.26 The Properties of Matter

(section shortcode: C10016 )

Let us now look at what we have learned about chemical bonds, intermolecular forces and the kinetic theory of

matter, and see whether this can help us to understand some of the macroscopic properties of materials.

1. Melting point

Definition: Melting point

The temperature at which a solid changes its phase or state to become a liquid. The

process is called melting and the reverse process (change in phase from liquid to solid) is

called freezing.

In order for a solid to melt, the energy of the particles must increase enough to overcome the bonds that are

holding the particles together. It makes sense then that a solid which is held together by strong bonds will

have a higher melting point than one where the bonds are weak, because more energy (heat) is needed

to break the bonds. In the examples we have looked at metals, ionic solids and some atomic lattices (e.g.

diamond) have high melting points, whereas the melting points for molecular solids and other atomic lattices

(e.g. graphite) are much lower. Generally, the intermolecular forces between molecular solids are weaker

than those between ionic and metallic solids.

2. Boiling point

Definition: Boiling point

The temperature at which a liquid changes its phase to become a gas. The process is

called evaporation and the reverse process is called condensation

When the temperature of a liquid increases, the average kinetic energy of the particles also increases

and they are able to overcome the bonding forces that are holding them in the liquid. When boiling point is

reached, evaporation takes place and some particles in the liquid become a gas. In other words, the energy

of the particles is too great for them to be held in a liquid anymore. The stronger the bonds within a liquid,

the higher the boiling point needs to be in order to break these bonds. Metallic and ionic compounds have

high boiling points while the boiling point for molecular liquids is lower. The data in Table 3.3 below may

help you to understand some of the concepts we have explained. Not all of the substances in the table are

solids at room temperature, so for now, let’s just focus on the boiling points for each of these substances.

What do you notice?

Substance Melting point (

C) Boiling point (

C)

Ethanol (C2H6O) - 114,3 78,4

Water 0 100

Mercury -38,83 356,73

Sodium chloride 801 1465

Table 3.3: The melting and boiling points for a number of substances

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CHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY1.26. THE PROPERTIES OF MATTER

You will have seen that substances such as ethanol, with relatively weak intermolecular forces, have the

lowest boiling point, while substances with stronger intermolecular forces such as sodium chloride and

mercury, must be heated much more if the particles are to have enough energy to overcome the forces that

are holding them together in the liquid. See the section (Section 2.5.1: Exercise: Forces and boiling point )

below for a further exercise on boiling point.

3. Density and viscosity The density of a solid is generally higher than that of a liquid because the

particles are held much more closely together and therefore there are more particles packed to-

gether in a particular volume. In other words, there is a greater mass of the substance in a partic-

ular volume. In general, density increases as the strength of the intermolecular forces increases.

Definition: Density

Density is a measure of the mass of a substance per unit volume.

Definition: Viscosity

Viscosity is a measure of how resistant a liquid is to flowing (in other words, how easy it is

to pour the liquid from one container to another).

Viscosity is also sometimes described as the ’thickness’ of a fluid. Think for example of syrup and how

slowly it pours from one container into another. Now compare this to how easy it is to pour water. The

viscosity of syrup is greater than the viscosity of water. Once again, the stronger the intermolecular forces

in the liquid, the greater its viscosity.

It should be clear now that we can explain a lot of the macroscopic properties of matter (i.e. the characteristics

we can see or observe) by understanding their microscopic structure and the way in which the atoms and

molecules that make up matter are held together.

1.26.1 Exercise: Forces and boiling point

The table below gives the molecular formula and the boiling point for a number of organic compounds called

alkanes (more on these compounds in grade 12). Refer to the table and then answer the questions that follow.

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1.26. THE PROPERTIES OF MATTERCHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY

Organic compound Molecular formula Boiling point (

C)

Methane CH4 -161.6

Ethane C2H6 - 88.6

Propane C3H8 -45

Butane C4H10 -0.5

Pentane C5H12 36.1

Hexane C6H14 69

Heptane C7H16 98.42

Octane C8H18 125.52

Table 3.4

Data from: http://www.wikipedia.com

1. Draw a graph to show the relationship between the number of carbon atoms in each alkane and its boiling

point. (Number of carbon atoms will go on the x-axis and boiling point on the y-axis).

2. Describe what you see.

3. Suggest a reason for what you have observed.

4. Why was it enough for us to use ’number of carbon atoms’ as a measure of the molecular weight of the

molecules?

Find the answers with the shortcodes: (1.) liP

1.26.2 Investigation : Determining the density of liquids:

Density is a very important property because it helps us to identify different materials. Every material, depending

on the elements that make it up and the arrangement of its atoms, will have a different density.

The equation for density is:

Density =Mass

Volume(3.1)

Discussion questions:

To calculate the density of liquids and solids, we need to be able to first determine their mass and volume. As a

group, think about the following questions:

• How would you determine the mass of a liquid?

• How would you determine the volume of an irregular solid?

Apparatus:

Laboratory mass balance, 10 ml and 100 ml graduated cylinders, thread, distilled water, two different liquids.

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CHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY1.26. THE PROPERTIES OF MATTER

Method:

Determine the density of the distilled water and two liquids as follows:

1. Measure and record the mass of a 10 ml graduated cyclinder.

2. Pour an amount of distilled water into the cylinder.

3. Measure and record the combined mass of the water and cylinder.

4. Record the volume of distilled water in the cylinder

5. Empty, clean and dry the graduated cylinder.

6. Repeat the above steps for the other two liquids you have.

7. Complete the table below.

Liquid Mass (g) Volume (ml) Density (g ·ml−1)

Distilled water

Liquid 1

Liquid 2

Table 3.5

1.26.3 Investigation : Determining the density of irregular solids:

Apparatus:

Use the same materials and equpiment as before (for the liquids). Also find a number of solids that have an

irregular shape.

Method:

Determine the density of irregular solids as follows:

1. Measure and record the mass of one of the irregular solids.

2. Tie a piece of thread around the solid.

3. Pour some water into a 100 ml graduated cylinder and record the volume.

4. Gently lower the solid into the water, keeping hold of the thread. Record the combined volume of the solid

and the water.

5. Determine the volume of the solid by subtracting the combined volume from the original volume of the water

only.

6. Repeat these steps for the second object.

7. Complete the table below.

Solid Mass (g) Volume (ml) Density (g ·ml−1)

Solid 1

Solid 2

Solid 3

Table 3.6

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1.27. SUMMARY CHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY

The following presentation provides a summary of the work covered in this chapter.

(Presentation: P10017 )

1.27 Summary

(section shortcode: C10018 )

• There are three states of matter: solid, liquid and gas.

• Diffusion is the movement of particles from a high concentration to a low concentration. Brownian motion

is the diffusion of many particles.

• The kinetic theory of matter attempts to explain the behaviour of matter in different phases.

• The kinetic theory of matter says that all matter is composed of particles which have a certain amount

of energy which allows them to move at different speeds depending on the temperature (energy). There

are spaces between the particles and also attractive forces between particles when they come close

together.

• Intramolecular force is the force between the atoms of a molecule, which holds them together. Intermolec-

ular force is a force between molecules, which holds them together.

• Understanding chemical bonds, intermolecular forces and the kinetic theory of matter can help to explain

many of the macroscopic properties of matter.

• Melting point is the temperature at which a solid changes its phase to become a liquid. The reverse

process (change in phase from liquid to solid) is called freezing. The stronger the chemical bonds and

intermolecular forces in a substance, the higher the melting point will be.

• Boiling point is the temperature at which a liquid changes phase to become a gas. The reverse pro-

cess (change in phase from gas to liquid) is called condensing. The stronger the chemical bonds and

intermolecular forces in a substance, the higher the boiling point will be.

• Density is a measure of the mass of a substance per unit volume.

• Viscosity is a measure of how resistant a liquid is to flowing.

1.28 End of chapter exercises

(section shortcode: C10019 )

1. Give one word or term for each of the following descriptions.

a. The property that determines how easily a liquid flows.

b. The change in phase from liquid to gas.

2. If one substance A has a melting point that is lower than the melting point of substance B, this suggests

that...

a. A will be a liquid at room temperature.

b. The chemical bonds in substance A are weaker than those in substance B.

c. The chemical bonds in substance A are stronger than those in substance B.

d. B will be a gas at room temperature.

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CHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY1.28. END OF CHAPTER EXERCISES

3. Boiling point is an important concept to understand.

a. Define ’boiling point’.

b. What change in phase takes place when a liquid reaches its boiling point?

c. What is the boiling point of water?

d. Use the kinetic theory of matter and your knowledge of intermolecular forces to explain why water

changes phase at this temperature.

4. Describe a solid in terms of the kinetic molecular theory.

5. Refer to the table below which gives the melting and boiling points of a number of elements and then answer

the questions that follow. (Data from http://www.chemicalelements.com)

Element Melting point Boiling point (

C)

copper 1083 2567

magnesium 650 1107

oxygen -218,4 -183

carbon 3500 4827

helium -272 -268,6

sulphur 112,8 444,6

Table 3.7

a. What state of matter (i.e. solid, liquid or gas) will each of these elements be in at room temperature?

b. Which of these elements has the strongest forces between its atoms? Give a reason for your answer.

c. Which of these elements has the weakest forces between its atoms? Give a reason for your answer.

Find the answers with the shortcodes:

(1.) l2t (2.) lip (3.) lim (4.) lgf (5.) liy

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1.28. END OF CHAPTER EXERCISESCHAPTER 1. STATES OF MATTER AND THE KINETIC MOLECULAR THEORY

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The Atom

1.29 Introduction

(section shortcode: C10020 )

The following video covers some of the properties of an atom.

Veritasium video on the atom - 1 (Video: P10021 )

We have now looked at many examples of the types of matter and materials that exist around us and we have

investigated some of the ways that materials are classified. But what is it that makes up these materials? And

what makes one material different from another? In order to understand this, we need to take a closer look at the

building block of matter - the atom. Atoms are the basis of all the structures and organisms in the universe. The

planets, sun, grass, trees, air we breathe and people are all made up of different combinations of atoms.

1.30 Project: Models of the atom

(section shortcode: C10022 )

Our current understanding of the atom came about over a long period of time, with many different people playing

a role. Conduct some research into the development of the different ideas of the atom and the people who

contributed to it. Some suggested people to look at are: JJ Thomson, Ernest Rutherford, Marie Curie, JC

Maxwell, Max Planck, Albert Einstein, Niels Bohr, Lucretius, LV de Broglie, CJ Davisson, LH Germer, Chadwick,

Werner Heisenberg, Max Born, Erwin Schrodinger, John Dalton, Empedocles, Leucippus, Democritus, Epicurus,

Zosimos, Maria the Jewess, Geber, Rhazes, Robert Boyle, Henry Cavendish, A Lavoisier and H Becquerel. You

do not need to find information on all these people, but try to find information about as many of them as possible.

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1.31. MODELS OF THE ATOM CHAPTER 1. THE ATOM

Make a list of the key contributions to a model of the atom that each of these people made and then make a

timeline of this information. (You can use an online tool such as Dipity1 to make a timeline.) Try to get a feel for

how it all eventually fit together into the modern understanding of the atom.

1.31 Models of the Atom

(section shortcode: C10023 )

It is important to realise that a lot of what we know about the structure of atoms has been developed over a long

period of time. This is often how scientific knowledge develops, with one person building on the ideas of someone

else. We are going to look at how our modern understanding of the atom has evolved over time.

The idea of atoms was invented by two Greek philosophers, Democritus and Leucippus in the fifth century BC.

The Greek word ατoµoν (atom) means indivisible because they believed that atoms could not be broken into

smaller pieces.

Nowadays, we know that atoms are made up of a positively charged nucleus in the centre surrounded by

negatively charged electrons. However, in the past, before the structure of the atom was properly understood,

scientists came up with lots of different models or pictures to describe what atoms look like.

Definition: Model

A model is a representation of a system in the real world. Models help us to understand

systems and their properties. For example, an atomic model represents what the structure

of an atom could look like, based on what we know about how atoms behave. It is not

necessarily a true picture of the exact structure of an atom.

1.31.1 The Plum Pudding Model

After the electron was discovered by J.J. Thomson in 1897, people realised that atoms were made up of even

smaller particles than they had previously thought. However, the atomic nucleus had not been discovered yet and

so the ’plum pudding model’ was put forward in 1904. In this model, the atom is made up of negative electrons

that float in a soup of positive charge, much like plums in a pudding or raisins in a fruit cake (Figure 4.2). In 1906,

Thomson was awarded the Nobel Prize for his work in this field. However, even with the Plum Pudding Model,

there was still no understanding of how these electrons in the atom were arranged.

1http://www.dipity.com/

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Figure 4.2: A schematic diagram to show what the atom looks like according to the Plum Pudding model

The discovery of radiation was the next step along the path to building an accurate picture of atomic structure. In

the early twentieth century, Marie Curie and her husband Pierre, discovered that some elements (the radioactive

elements) emit particles, which are able to pass through matter in a similar way to X-rays (read more about this

in Grade 11). It was Ernest Rutherford who, in 1911, used this discovery to revise the model of the atom.

1.31.2 Rutherford’s model of the atom

Radioactive elements emit different types of particles. Some of these are positively charged alpha (α) particles.

Rutherford carried out a series of experiments where he bombarded sheets of gold foil with these particles, to try

to get a better understanding of where the positive charge in the atom was. A simplified diagram of his experiment

is shown in Figure 4.3.

Figure 4.3: Rutherford’s gold foil experiment. Figure (a) shows the path of the α particles after they hit the gold

sheet. Figure (b) shows the arrangement of atoms in the gold sheets and the path of the α particles in relation

to this.

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1.31. MODELS OF THE ATOM CHAPTER 1. THE ATOM

Rutherford set up his experiment so that a beam of alpha particles was directed at the gold sheets. Behind

the gold sheets was a screen made of zinc sulphide. This screen allowed Rutherford to see where the alpha

particles were landing. Rutherford knew that the electrons in the gold atoms would not really affect the path of

the alpha particles, because the mass of an electron is so much smaller than that of a proton. He reasoned that

the positively charged protons would be the ones to repel the positively charged alpha particles and alter their

path.

What he discovered was that most of the alpha particles passed through the foil undisturbed and could be

detected on the screen directly behind the foil (A). Some of the particles ended up being slightly deflected onto

other parts of the screen (B). But what was even more interesting was that some of the particles were deflected

straight back in the direction from where they had come (C)! These were the particles that had been repelled by

the positive protons in the gold atoms. If the Plum Pudding model of the atom were true then Rutherford would

have expected much more repulsion, since the positive charge according to that model is distributed throughout

the atom. But this was not the case. The fact that most particles passed straight through suggested that the

positive charge was concentrated in one part of the atom only.

Rutherford’s work led to a change in ideas around the atom. His new model described the atom as a tiny, dense,

positively charged core called a nucleus surrounded by lighter, negatively charged electrons. Another way of

thinking about this model was that the atom was seen to be like a mini solar system where the electrons orbit

the nucleus like planets orbiting around the sun. A simplified picture of this is shown in Figure 4.4. This model is

sometimes known as the planetary model of the atom.

Figure 4.4: Rutherford’s model of the atom

1.31.3 The Bohr Model

There were, however, some problems with this model: for example it could not explain the very interesting

observation that atoms only emit light at certain wavelengths or frequencies. Niels Bohr solved this problem by

proposing that the electrons could only orbit the nucleus in certain special orbits at different energy levels around

the nucleus. The exact energies of the orbitals in each energy level depends on the type of atom. Helium for

example, has different energy levels to Carbon. If an electron jumps down from a higher energy level to a lower

energy level, then light is emitted from the atom. The energy of the light emitted is the same as the gap in

the energy between the two energy levels. You can read more about this in "Energy quantisation and electron

configuration" (Section 3.8: Electron configuration). The distance between the nucleus and the electron in the

lowest energy level of a hydrogen atom is known as the Bohr radius.

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NOTE: Light has the properties of both a particle and a wave! Einstein discovered

that light comes in energy packets which are called photons. When an electron in an

atom changes energy levels, a photon of light is emitted. This photon has the same

energy as the difference between the two electron energy levels.

1.31.4 Other models of the atom

Although the most common model of the atom is the Bohr model, scientists have not stopped thinking about other

ways to describe atoms. One of the most important contributions to atomic theory (the field of science that looks

at atoms) was the development of quantum theory. Schrodinger, Heisenberg, Born and many others have had

a role in developing quantum theory. The description of an atom by quantum theory is very complex and is only

covered at university level.

1.31.5 Models of the atom

Match the information in column A, with the key discoverer in column B.

Column A Column B

Discovery of electrons and the plum pudding model Niels Bohr

Arrangement of electrons Marie Curie and her husband, Pierre

Atoms as the smallest building block of matter Ancient Greeks

Discovery of the nucleus JJ Thomson

Discovery of radiation Rutherford

Table 4.1

Find the answers with the shortcodes:

(1.) l4g

1.32 The size of atoms

(section shortcode: C10024 )

It is difficult sometimes to imagine the size of an atom, or its mass, because we cannot see an atom and also

because we are not used to working with such small measurements.

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1.32.1 How heavy is an atom?

It is possible to determine the mass of a single atom in kilograms. But to do this, you would need very modern

mass spectrometers and the values you would get would be very clumsy and difficult to use. The mass of a carbon

atom, for example, is about 1, 99 × 10−26 kg, while the mass of an atom of hydrogen is about 1, 67 × 10−27 kg.

Looking at these very small numbers makes it difficult to compare how much bigger the mass of one atom is

when compared to another.

To make the situation simpler, scientists use a different unit of mass when they are describing the mass of an

atom. This unit is called the atomic mass unit (amu). We can abbreviate (shorten) this unit to just ’u’. Scientists

use the carbon standard to determine amu. The carbon standard assigns carbon an atomic mass of 12 u. Using

the carbon standard the mass of an atom of hydrogen will be 1 u. You can check this by dividing the mass of

a carbon atom in kilograms (see above) by the mass of a hydrogen atom in kilograms (you will need to use a

calculator for this!). If you do this calculation, you will see that the mass of a carbon atom is twelve times greater

than the mass of a hydrogen atom. When we use atomic mass units instead of kilograms, it becomes easier to

see this. Atomic mass units are therefore not giving us the actual mass of an atom, but rather its mass relative to

the mass of one (carefully chosen) atom in the Periodic Table. Although carbon is the usual element to compare

other elements to, oxygen and hydrogen have also been used. The important thing to remember here is that the

atomic mass unit is relative to one (carefully chosen) element. The atomic masses of some elements are shown

in the table (Table 4.2) below.

Element Atomic mass (u)

Carbon (C) 12

Nitrogen (N) 14

Bromine (Br) 80

Magnesium (Mg) 24

Potassium (K) 39

Calcium (Ca) 40

Oxygen (O) 16

Table 4.2: The atomic mass number of some of the elements

The actual value of 1 atomic mass unit is 1, 67× 10−24 g or 1, 67× 10−27 kg. This is a very tiny mass!

1.32.2 How big is an atom?

TIP: pm stands for picometres. 1 pm = 10−12 m

Atomic radius also varies depending on the element. On average, the radius of an atom ranges from 32 pm(Helium) to 225 pm (Caesium). Using different units, 100 pm = 1Angstrom, and 1 Angstrom = 10−10 m. That

is the same as saying that 1 Angstrom = 0, 0000000010 m or that 100 pm = 0, 0000000010 m! In other words,

the diameter of an atom ranges from 0, 0000000010m to 0, 0000000067m. This is very small indeed.

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The atomic radii given above are for the whole atom (nucleus and electrons). The nucleus itself is even smaller

than this by a factor of about 23 000 in uranium and 145 000 in hydrogen. If the nucleus were the size of a golf

ball, then the nearest electrons would be about one kilometer away! This should give help you realise that the

atom is mostly made up of empty space.

1.33 Atomic structure

(section shortcode: C10025 )

As a result of the work done by previous scientists on atomic models (that we discussed in "Models of the Atom"

(Section 3.3: Models of the Atom)), scientists now have a good idea of what an atom looks like. This knowledge

is important because it helps us to understand why materials have different properties and why some materials

bond with others. Let us now take a closer look at the microscopic structure of the atom.

So far, we have discussed that atoms are made up of a positively charged nucleus surrounded by one or more

negatively charged electrons. These electrons orbit the nucleus.

1.33.1 The Electron

The electron is a very light particle. It has a mass of 9, 11× 10−31 kg. Scientists believe that the electron can be

treated as a point particle or elementary particle meaning that it can’t be broken down into anything smaller.

The electron also carries one unit of negative electric charge which is the same as 1, 6× 10−19 C (Coulombs).

The electrons determine the charge on an atom. If the number of electrons is the same as the number of protons

then the atom will be neutral. If the number of electrons is greater than the number of protons then the atom

will be negatively charged. If the number of electrons is less than the number of protons then the atom will be

positively charged. Atoms that are not neutral are called ions. Ions will be covered in more detail in a later

chapter. For now all you need to know is that for each electron you remove from an atom you loose −1 of charge

and for each electron that you add to an atom you gain +1 of charge. For example, the charge on an atom of

sodium after removing one electron is −1.

1.33.2 The Nucleus

Unlike the electron, the nucleus can be broken up into smaller building blocks called protons and neutrons.

Together, the protons and neutrons are called nucleons.

The Proton

Each proton carries one unit of positive electric charge. Since we know that atoms are electrically neutral,

i.e. do not carry any extra charge, then the number of protons in an atom has to be the same as the number of

electrons to balance out the positive and negative charge to zero. The total positive charge of a nucleus is equal

to the number of protons in the nucleus. The proton is much heavier than the electron (10 000 times heavier!)

and has a mass of 1, 6726× 10−27 kg. When we talk about the atomic mass of an atom, we are mostly referring

to the combined mass of the protons and neutrons, i.e. the nucleons.

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The Neutron

The neutron is electrically neutral i.e. it carries no charge at all. Like the proton, it is much heavier than the

electron and its mass is 1, 6749× 10−27 kg (slightly heavier than the proton).

NOTE: Rutherford predicted (in 1920) that another kind of particle must be present

in the nucleus along with the proton. He predicted this because if there were only

positively charged protons in the nucleus, then it should break into bits because of

the repulsive forces between the like-charged protons! Also, if protons were the only

particles in the nucleus, then a helium nucleus (atomic number 2) would have two pro-

tons and therefore only twice the mass of hydrogen. However, it is actually four times

heavier than hydrogen. This suggested that there must be something else inside the

nucleus as well as the protons. To make sure that the atom stays electrically neutral,

this particle would have to be neutral itself. In 1932 James Chadwick discovered the

neutron and measured its mass.

proton neutron electron

Mass (kg) 1, 6726× 10−27 1, 6749× 10−27 9, 11× 10−31

Units of charge +1 0 −1

Charge (C) 1, 6× 10−19 0 −1, 6× 10−19

Table 4.3: Summary of the particles inside the atom

NOTE: Unlike the electron which is thought to be a point particle and unable to be

broken up into smaller pieces, the proton and neutron can be divided. Protons and

neutrons are built up of smaller particles called quarks. The proton and neutron are

made up of 3 quarks each.

1.34 Atomic number and atomic mass number

(section shortcode: C10026 )

The chemical properties of an element are determined by the charge of its nucleus, i.e. by the number of

protons. This number is called the atomic number and is denoted by the letter Z.

Definition: Atomic number (Z)

The number of protons in an atom

You can find the atomic number on the periodic table. The atomic number is an integer and ranges from 1 to

about 118.

The mass of an atom depends on how many nucleons its nucleus contains. The number of nucleons, i.e. the

total number of protons plus neutrons, is called the atomic mass number and is denoted by the letter A.

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Definition: Atomic mass number (A)

The number of protons and neutrons in the nucleus of an atom

Standard notation shows the chemical symbol, the atomic mass number and the atomic number of an element

as follows:

NOTE: A nuclide is a distinct kind of atom or nucleus characterized by the number of

protons and neutrons in the atom. To be absolutely correct, when we represent atoms

like we do here, then we should call them nuclides.

For example, the iron nucleus which has 26 protons and 30 neutrons, is denoted as:

5626Fe (4.1)

where the atomic number is Z = 26 and the mass number A = 56. The number of neutrons is simply the

difference N = A− Z.

TIP: Don’t confuse the notation we have used above with the way this information

appears on the Periodic Table. On the Periodic Table, the atomic number usually

appears in the top lefthand corner of the block or immediately above the element’s

symbol. The number below the element’s symbol is its relative atomic mass. This is

not exactly the same as the atomic mass number. This will be explained in "Isotopes"

(Section 3.7: Isotopes). The example of iron is shown below.

You will notice in the example of iron that the atomic mass number is more or less the same as its atomic mass.

Generally, an atom that contains n nucleons (protons and neutrons), will have a mass approximately equal to

nu. For example the mass of a C − 12 atom which has 6 protons, 6 neutrons and 6 electrons is 12u, where the

protons and neutrons have about the same mass and the electron mass is negligible.

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Exercise 4.1 Use standard notation to represent sodium and give the

number of protons, neutrons and electrons in the element.

Solution to Exercise

Step 1. Sodium is given by NaStep 2. Sodium has 11 protons, so we have: 11NaStep 3. Sodium has 12 neutrons.

Step 4. A = N + Z = 12 + 11 = 23Step 5. In standard notation sodium is given by: 23

11Na. The number of

protons is 11, the number of neutrons is 12 and the number of

electrons is 11.

1.34.1 The structure of the atom

1. Explain the meaning of each of the following terms:

a. nucleus

b. electron

c. atomic mass

2. Complete the following table: (Note: You will see that the atomic masses on the Periodic Table are not

whole numbers. This will be explained later. For now, you can round off to the nearest whole number.)

Element Atomic mass Atomic num-

ber

Number of

protons

Number of

electrons

Number of

neutrons

Mg 24 12

O 8

17

Ni 28

40 20

Zn

0

continued on next page

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CHAPTER 1. THE ATOM 1.35. ISOTOPES

C 12 6

Table 4.4

3. Use standard notation to represent the following elements:

a. potassium

b. copper

c. chlorine

4. For the element 3517Cl, give the number of ...

a. protons

b. neutrons

c. electrons

... in the atom.

5. Which of the following atoms has 7 electrons?

a. 52He

b. 136 C

c. 73Li

d. 157 N

6. In each of the following cases, give the number or the element symbol represented by ’X’.

a. 4018X

b. x20Ca

c. 31x P

7. Complete the following table:

A Z N

23592 U

23892 U

Table 4.5

In these two different forms of Uranium...

a. What is the same?

b. What is different?

Uranium can occur in different forms, called isotopes. You will learn more about isotopes in "Isotopes"

(Section 3.7: Isotopes).

Find the answers with the shortcodes:

(1.) ll0 (2.) ll8 (3.) ll9 (4.) llX (5.) llk (6.) llK

(7.) llB

1.35 Isotopes

(section shortcode: C10027 )

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1.35.1 What is an isotope?

The chemical properties of an element depend on the number of protons and electrons inside the atom. So if a

neutron or two is added or removed from the nucleus, then the chemical properties will not change. This means

that such an atom would remain in the same place in the Periodic Table. For example, no matter how many

neutrons we add or subtract from a nucleus with 6 protons, that element will always be called carbon and have

the element symbol C (see the Table of Elements). Atoms which have the same number of protons, but a different

number of neutrons, are called isotopes.

Definition: Isotope

The isotope of a particular element is made up of atoms which have the same number of

protons as the atoms in the original element, but a different number of neutrons.

The different isotopes of an element have the same atomic number Z but different mass numbers A because

they have a different number of neutrons N . The chemical properties of the different isotopes of an element are

the same, but they might vary in how stable their nucleus is. Note that we can also write elements as X − Awhere the X is the element symbol and the A is the atomic mass of that element. For example, C−12 has an

atomic mass of 12 and Cl−35 has an atomic mass of 35 u, while Cl−37 has an atomic mass of 37 u.

NOTE: In Greek, “same place” reads as‘ι σoςτ

‘o πoς (isos topos). This is why atoms

which have the same number of protons, but different numbers of neutrons, are called

isotopes. They are in the same place on the Periodic Table!

The following worked examples will help you to understand the concept of an isotope better.

Exercise 4.2: Isotopes For the element 23492 U (uranium), use standard

notation to describe:

1. the isotope with 2 fewer neutrons

2. the isotope with 4 more neutrons

Solution to Exercise

Step 1. We know that isotopes of any element have the same number of

protons (same atomic number) in each atom, which means that

they have the same chemical symbol. However, they have a differ-

ent number of neutrons, and therefore a different mass number.

Step 2. Therefore, any isotope of uranium will have the symbol:

U (4.2)

Also, since the number of protons in uranium isotopes is always

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CHAPTER 1. THE ATOM 1.35. ISOTOPES

the same, we can write down the atomic number:

92U (4.3)

Now, if the isotope we want has 2 fewer neutrons than 23492 U, then

we take the original mass number and subtract 2, which gives:

23292 U (4.4)

Following the steps above, we can write the isotope with 4 more

neutrons as:

23892 U (4.5)

Exercise 4.3: Isotopes Which of the following are isotopes of 4020Ca?

• 4019K

• 4220Ca

• 4018Ar

Solution to Exercise

Step 1. We know that isotopes have the same atomic number but different

mass numbers.

Step 2. You need to look for the element that has the same atomic number

but a different atomic mass number. The only element is 4220Ca.

What is different is that there are 2 more neutrons than in the orig-

inal element.

Exercise 4.4: Isotopes For the sulphur isotope 3316S, give the number

of...

a. protons

b. nucleons

c. electrons

d. neutrons

Solution to Exercise

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1.35. ISOTOPES CHAPTER 1. THE ATOM

Step 1. Z = 16, therefore the number of protons is 16 (answer to (a)).

Step 2. A = 33, therefore the number of nucleons is 33 (answer to (b)).

Step 3. The atom is neutral, and therefore the number of electrons is the

same as the number of protons. The number of electrons is 16

(answer to (c)).

Step 4.

N = A− Z = 33− 16 = 17 (4.6)

The number of neutrons is 17 (answer to (d)).

Isotopes

1. Atom A has 5 protons and 5 neutrons, and atom B has 6 protons and 5 neutrons. These atoms are...

a. allotropes

b. isotopes

c. isomers

d. atoms of different elements

2. For the sulphur isotopes, 3216S and 34

16S, give the number of...

a. protons

b. nucleons

c. electrons

d. neutrons

3. Which of the following are isotopes of 3517Cl?

a. 1735Cl

b. 3517Cl

c. 3717Cl

4. Which of the following are isotopes of U−235? (X represents an element symbol)

a. 23892 X

b. 23890 X

c. 23592 X

Find the answers with the shortcodes:

(1.) ll4 (2.) llZ (3.) llW (4.) llD

1.35.2 Relative atomic mass

It is important to realise that the atomic mass of isotopes of the same element will be different because they

have a different number of nucleons. Chlorine, for example, has two common isotopes which are chlorine-35 and

chlorine-37. Chlorine-35 has an atomic mass of 35 u, while chlorine-37 has an atomic mass of 37 u. In the world

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CHAPTER 1. THE ATOM 1.35. ISOTOPES

around us, both of these isotopes occur naturally. It doesn’t make sense to say that the element chlorine has an

atomic mass of 35 u, or that it has an atomic mass of 37 u. Neither of these are absolutely true since the mass

varies depending on the form in which the element occurs. We need to look at how much more common one is

than the other in order to calculate the relative atomic mass for the element chlorine. This is the number that

you find on the Periodic Table.

Definition: Relative atomic mass

Relative atomic mass is the average mass of one atom of all the naturally occurring iso-

topes of a particular chemical element, expressed in atomic mass units.

NOTE: The relative atomic mass of some elements depends on where on Earth the el-

ement is found. This is because the isotopes can be found in varying ratios depending

on certain factors such as geological composition, etc. The International Union of Pure

and Applied Chemistry (IUPAC) has decided to give the relative atomic mass of some

elements as a range to better represent the varying isotope ratios on the Earth. For

the calculations that you will do at high school, it is enough to simply use one number

without worrying about these ranges.

Exercise 4.5: The relative atomic mass of an isotopic element The

element chlorine has two isotopes, chlorine-35 and chlorine-37. The

abundance of these isotopes when they occur naturally is 75% chlorine-

35 and 25% chlorine-37. Calculate the average relative atomic mass for

chlorine.

Solution to Exercise

Step 1. Contribution of Cl−35 =(

75100 × 35

)

= 26, 25 u

Step 2. Contribution of Cl−37 =(

25100 × 37

)

= 9, 25 uStep 3. Relative atomic mass of chlorine = 26, 25 u + 9, 25 u = 35, 5 u

If you look on the periodic table, the average relative atomic mass

for chlorine is 35, 5 u. You will notice that for many elements, the

relative atomic mass that is shown is not a whole number. You

should now understand that this number is the average relative

atomic mass for those elements that have naturally occurring iso-

topes.

This simulation allows you to see how isotopes and relative atomic mass are inter related.

(Simulation: lbT)

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1.35. ISOTOPES CHAPTER 1. THE ATOM

Isotopes

1. Complete the table below:

Isotope Z A Protons Neutrons Electrons

Carbon-12

Carbon-14

Chlorine-35

Chlorine-37

Table 4.6

2. If a sample contains 90% carbon-12 and 10% carbon-14, calculate the relative atomic mass of an atom in

that sample.

3. If a sample contains 22,5% Cl-37 and 77,5% Cl-35, calculate the relative atomic mass of an atom in that

sample.

Find the answers with the shortcodes:

(1.) llj (2.) llb (3.) llT

Group Discussion : The changing nature of scientific knowledge

Scientific knowledge is not static: it changes and evolves over time as scientists build on the ideas of others to

come up with revised (and often improved) theories and ideas. In this chapter for example, we saw how peoples’

understanding of atomic structure changed as more information was gathered about the atom. There are many

more examples like this one in the field of science. For example, think about our knowledge of the solar system

and the origin of the universe, or about the particle and wave nature of light.

Often, these changes in scientific thinking can be very controversial because they disturb what people have come

to know and accept. It is important that we realise that what we know now about science may also change. An

important part of being a scientist is to be a critical thinker. This means that you need to question information that

you are given and decide whether it is accurate and whether it can be accepted as true. At the same time, you

need to learn to be open to new ideas and not to become stuck in what you believe is right... there might just be

something new waiting around the corner that you have not thought about!

In groups of 4-5, discuss the following questions:

• Think about some other examples where scientific knowledge has changed because of new ideas and

discoveries:

· What were these new ideas?

· Were they controversial? If so, why?

· What role (if any) did technology play in developing these new ideas?

· How have these ideas affected the way we understand the world?

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CHAPTER 1. THE ATOM 1.36. ELECTRON CONFIGURATION

• Many people come up with their own ideas about how the world works. The same is true in science. So

how do we, and other scientists, know what to believe and what not to? How do we know when new ideas

are ’good’ science or ’bad’ science? In your groups, discuss some of the things that would need to be done

to check whether a new idea or theory was worth listening to, or whether it was not.

• Present your ideas to the rest of the class.

1.36 Electron configuration

(section shortcode: C10028 )

1.36.1 The energy of electrons

You will remember from our earlier discussions that an atom is made up of a central nucleus, which contains

protons and neutrons and that this nucleus is surrounded by electrons. Although these electrons all have the

same charge and the same mass, each electron in an atom has a different amount of energy. Electrons that

have the lowest energy are found closest to the nucleus where the attractive force of the positively charged

nucleus is the greatest. Those electrons that have higher energy, and which are able to overcome the attractive

force of the nucleus, are found further away.

1.36.2 Energy quantisation and line emission spectra (Not in CAPS, included for com-

pleteness)

If the energy of an atom is increased (for example when a substance is heated), the energy of the electrons

inside the atom can be increased (when an electron has a higher energy than normal it is said to be "excited").

For the excited electron to go back to its original energy (called the ground state), it needs to release energy. It

releases energy by emitting light. If one heats up different elements, one will see that for each element, light is

emitted only at certain frequencies (or wavelengths). Instead of a smooth continuum of frequencies, we see lines

(called emission lines) at particular frequencies. These frequencies correspond to the energy of the emitted light.

If electrons could be excited to any energy and lose any amount of energy, there would be a continuous spread

of light frequencies emitted. However, the sharp lines we see mean that there are only certain particular energies

that an electron can be excited to, or can lose, for each element.

You can think of this like going up a flight of steps: you can’t lift your foot by any amount to go from the ground

to the first step. If you lift your foot too low you’ll bump into the step and be stuck on the ground level. You have

to lift your foot just the right amount (the height of the step) to go to the next step, and so on. The same goes for

electrons and the amount of energy they can have. This is called quantisation of energy because there are only

certain quantities of energy that an electron can have in an atom. Like steps, we can think of these quantities as

energy levels in the atom. The energy of the light released when an electron drops down from a higher energy

level to a lower energy level is the same as the difference in energy between the two levels.

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1.36.3 Electron configuration

We will start with a very simple view of the arrangement or configuration of electrons around an atom. This

view simply states that electrons are arranged in energy levels (or shells) around the nucleus of an atom. These

energy levels are numbered 1, 2, 3, etc. Electrons that are in the first energy level (energy level 1) are closest to

the nucleus and will have the lowest energy. Electrons further away from the nucleus will have a higher energy.

In the following examples, the energy levels are shown as concentric circles around the central nucleus. The

important thing to know for these diagrams is that the first energy level can hold 2 electrons, the second energy

level can hold 8 electrons and the third energy level can hold 8 electrons.

1. Lithium Lithium (Li) has an atomic number of 3, meaning that in a neutral atom, the number of electrons

will also be 3. The first two electrons are found in the first energy level, while the third electron is found in

the second energy level (Figure 4.8).

Figure 4.8: The arrangement of electrons in a lithium atom.

2. Fluorine Fluorine (F) has an atomic number of 9, meaning that a neutral atom also has 9 electrons. The

first 2 electrons are found in the first energy level, while the other 7 are found in the second energy level

(Figure 4.9).

Figure 4.9: The arrangement of electrons in a fluorine atom.

3. Argon Argon has an atomic number of 18, meaning that a neutral atom also has 18 electrons. The first 2

electrons are found in the first energy level, the next 8 are found in the second energy level, and the last 8

are found in the third energy level (Figure 4.10).

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Figure 4.10: The arrangement of electrons in an argon atom.

But the situation is slightly more complicated than this. Within each energy level, the electrons move in orbitals.

An orbital defines the spaces or regions where electrons move.

Definition: Atomic orbital

An atomic orbital is the region in which an electron may be found around a single atom.

There are different orbital shapes, but we will be mainly dealing with only two. These are the ’s’ and ’p’ orbitals

(there are also ’d’ and ’f’ orbitals). The ’s’ orbitals are spherical and the ’p’ orbitals are dumbbell shaped.

Figure 4.11: The shapes of orbitals. a) shows an ’s’ orbital, b) shows a single ’p’ orbital and c) shows the three

’p’ orbitals.

The first energy level contains only one ’s’ orbital, the second energy level contains one ’s’ orbital and three

’p’ orbitals and the third energy level contains one ’s’ orbital and three ’p’ orbitals (as well as 5 ’d’ orbitals).

Within each energy level, the ’s’ orbital is at a lower energy than the ’p’ orbitals. This arrangement is shown in

Figure 4.12.

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Figure 4.12: The positions of the first ten orbitals of an atom on an energy diagram. Note that each block is able

to hold two electrons.

This diagram also helps us when we are working out the electron configuration of an element. The electron

configuration of an element is the arrangement of the electrons in the shells and subshells. There are a few

guidelines for working out the electron configuration. These are:

• Each orbital can only hold two electrons. Electrons that occur together in an orbital are called an electron

pair.

• An electron will always try to enter an orbital with the lowest possible energy.

• An electron will occupy an orbital on its own, rather than share an orbital with another electron. An electron

would also rather occupy a lower energy orbital with another electron, before occupying a higher energy

orbital. In other words, within one energy level, electrons will fill an ’s’ orbital before starting to fill ’p’ orbitals.

• The s subshell can hold 2 electrons

• The p subshell can hold 6 electrons

In the examples you will cover, you will mainly be filling the s and p subshells. Occasionally you may get an

example that has the d subshell. The f subshell is more complex and is not covered at this level.

The way that electrons are arranged in an atom is called its electron configuration.

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Definition: Electron configuration

Electron configuration is the arrangement of electrons in an atom, molecule or other phys-

ical structure.

An element’s electron configuration can be represented using Aufbau diagrams or energy level diagrams. An

Aufbau diagram uses arrows to represent electrons. You can use the following steps to help you to draw an

Aufbau diagram:

1. Determine the number of electrons that the atom has.

2. Fill the ’s’ orbital in the first energy level (the 1s orbital) with the first two electrons.

3. Fill the ’s’ orbital in the second energy level (the 2s orbital) with the second two electrons.

4. Put one electron in each of the three ’p’ orbitals in the second energy level (the 2p orbitals) and then if there

are still electrons remaining, go back and place a second electron in each of the 2p orbitals to complete the

electron pairs.

5. Carry on in this way through each of the successive energy levels until all the electrons have been drawn.

TIP: When there are two electrons in an orbital, the electrons are called an electron

pair. If the orbital only has one electron, this electron is said to be an unpaired elec-

tron. Electron pairs are shown with arrows pointing in opposite directions. You may

hear people talking of the Pauli exclusion principle. This principle says that electrons

have a property known as spin and two electrons in an orbital will not spin the same

way. This is why we use arrows pointing in opposite directions. An arrow pointing up

denotes an electron spinning one way and an arrow pointing downwards denotes an

electron spinning the other way.

NOTE: Aufbau is the German word for ’building up’. Scientists used this term since this

is exactly what we are doing when we work out electron configuration, we are building

up the atoms structure.

Sometimes people refer to Hund’s rule for electron configuration. This rule simply says that electrons would rather

be in a subshell on it’s own then share a subshell. This is why, when you are filling the subshells you put one

electron in each subshell and only if there are extra electrons do you go back and fill the subshell, before moving

onto the next energy level.

An Aufbau diagram for the element Lithium is shown in Figure 4.13.

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Figure 4.13: The electron configuration of Lithium, shown on an Aufbau diagram

A special type of notation is used to show an atom’s electron configuration. The notation describes the energy

levels, orbitals and the number of electrons in each. For example, the electron configuration of lithium is 1s22s1.

The number and letter describe the energy level and orbital and the number above the orbital shows how many

electrons are in that orbital.

Aufbau diagrams for the elements fluorine and argon are shown in Figure 4.14 and Figure 4.15 respectively. Using

standard notation, the electron configuration of fluorine is 1s22s22p5 and the electron configuration of argon is

1s22s22p6.

Figure 4.14: An Aufbau diagram showing the electron configuration of fluorine

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Figure 4.15: An Aufbau diagram showing the electron configuration of argon

Exercise 4.6: Aufbau diagrams Give the electron configuration for

sodium (Na) and draw an aufbau diagram.

Solution to Exercise

Step 1. Sodium has 11 electrons.

Step 2. We start by placing two electrons in the 1s orbital: 1s2. Now we

have 9 electrons left to place in orbitals, so we put two in the 2sorbital: 2s2. There are now 7 electrons to place in orbitals so we

place 6 of them in the 2p orbital: 2p6. The last electron goes into

the 3s orbital: 3s1.

Step 3. The electron configuration is: 1s22s22p63s1

Step 4. Using the electron configuration we get the following diagram:

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1.36.4 Core and valence electrons

Electrons in the outermost energy level of an atom are called valence electrons. The electrons that are in the

energy shells closer to the nucleus are called core electrons. Core electrons are all the electrons in an atom,

excluding the valence electrons. An element that has its valence energy level full is more stable and less likely to

react than other elements with a valence energy level that is not full.

Definition: Valence electrons

The electrons in the outer energy level of an atom

Definition: Core electrons

All the electrons in an atom, excluding the valence electrons

1.36.5 The importance of understanding electron configuration

By this stage, you may well be wondering why it is important for you to understand how electrons are arranged

around the nucleus of an atom. Remember that during chemical reactions, when atoms come into contact with

one another, it is the electrons of these atoms that will interact first. More specifically, it is the valence electrons

of the atoms that will determine how they react with one another.

To take this a step further, an atom is at its most stable (and therefore unreactive) when all its orbitals are full. On

the other hand, an atom is least stable (and therefore most reactive) when its valence electron orbitals are not

full. This will make more sense when we go on to look at chemical bonding in a later chapter. To put it simply, the

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valence electrons are largely responsible for an element’s chemical behaviour and elements that have the same

number of valence electrons often have similar chemical properties.

One final point to note about electron configurations is stability. Which configurations are stable and which are

not? Very simply, the most stable configurations are the ones that have full energy levels. These configurations

occur in the noble gases. The noble gases are very stable elements that do not react easily (if at all) with any

other elements. This is due to the full energy levels. All elements would like to reach the most stable electron

configurations, i.e. all elements want to be noble gases. This principle of stability is sometimes referred to as the

octet rule. An octet is a set of 8, and the number of electrons in a full energy level is 8.

Experiment: Flame tests

Aim:

To determine what colour a metal cation will cause a flame to be.

Apparatus:

Watch glass, bunsen burner, methanol, bamboo sticks, metal salts (e.g. NaCl, CuCl2, CaCl2, KCl, etc. ) and

metal powders (e.g. copper, magnesium, zinc, iron, etc.)

Method:

For each salt or powder do the following:

1. Dip a clean bamboo stick into the methanol

2. Dip the stick into the salt or powder

3. Wave the stick through the flame from the bunsen burner. DO NOT hold the stick in the flame, but rather

wave it back and forth through the flame.

4. Observe what happens

Results:

Record your results in a table, listing the metal salt and the colour of the flame.

Conclusion:

You should have observed different colours for each of the metal salts and powders that you tested.

The above experiment on flame tests relates to the line emission spectra of the metals. These line emission

spectra are a direct result of the arrangement of the electrons in metals.

Energy diagrams and electrons

1. Draw Aufbau diagrams to show the electron configuration of each of the following elements:

a. magnesium

b. potassium

c. sulphur

d. neon

e. nitrogen

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1.36. ELECTRON CONFIGURATION CHAPTER 1. THE ATOM

2. Use the Aufbau diagrams you drew to help you complete the following table:

Element No. of energy

levels

No. of core elec-

trons

No. of valence

electrons

Electron config-

uration (stan-

dard notation)

Mg

K

S

Ne

N

Table 4.7

3. Rank the elements used above in order of increasing reactivity. Give reasons for the order you give.

Find the answers with the shortcodes: ll2

Group work : Building a model of an atom

Earlier in this chapter, we talked about different ’models’ of the atom. In science, one of the uses of models

is that they can help us to understand the structure of something that we can’t see. In the case of the atom,

models help us to build a picture in our heads of what the atom looks like.

Models are often simplified. The small toy cars that you may have played with as a child are models. They

give you a good idea of what a real car looks like, but they are much smaller and much simpler. A model

cannot always be absolutely accurate and it is important that we realise this so that we don’t build up a

false idea about something.

In groups of 4-5, you are going to build a model of an atom. Before you start, think about these questions:

• What information do I know about the structure of the atom? (e.g. what parts make it up? how big is

it?)

• What materials can I use to represent these parts of the atom as accurately as I can?

• How will I put all these different parts together in my model?

As a group, share your ideas and then plan how you will build your model. Once you have built your model,

discuss the following questions:

• Does our model give a good idea of what the atom actually looks like?

• In what ways is our model inaccurate? For example, we know that electrons move around the atom’s

nucleus, but in your model, it might not have been possible for you to show this.

• Are there any ways in which our model could be improved?

Now look at what other groups have done. Discuss the same questions for each of the models you see and

record your answers.

The following simulation allows you to build an atom

(Simulation: lbb)

This is another simulation that allows you to build an atom. This simulation also provides a summary of

what you have learnt so far.

(Simulation: lbj)

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CHAPTER 1. THE ATOM 1.37. SUMMARY

1.37 Summary

(section shortcode: C10029 )

• Much of what we know today about the atom, has been the result of the work of a number of scientists

who have added to each other’s work to give us a good understanding of atomic structure.

• Some of the important scientific contributors include J.J.Thomson (discovery of the electron, which

led to the Plum Pudding Model of the atom), Ernest Rutherford (discovery that positive charge is

concentrated in the centre of the atom) and Niels Bohr (the arrangement of electrons around the

nucleus in energy levels).

• Because of the very small mass of atoms, their mass is measured in atomic mass units (u). 1 u =1, 67× 10−24 g.

• An atom is made up of a central nucleus (containing protons and neutrons), surrounded by elec-

trons.

• The atomic number (Z) is the number of protons in an atom.

• The atomic mass number (A) is the number of protons and neutrons in the nucleus of an atom.

• The standard notation that is used to write an element, is AZX, where X is the element symbol, A is

the atomic mass number and Z is the atomic number.

• The isotope of a particular element is made up of atoms which have the same number of protons as

the atoms in the original element, but a different number of neutrons. This means that not all atoms of

an element will have the same atomic mass.

• The relative atomic mass of an element is the average mass of one atom of all the naturally occurring

isotopes of a particular chemical element, expressed in atomic mass units. The relative atomic mass

is written under the elements’ symbol on the Periodic Table.

• The energy of electrons in an atom is quantised. Electrons occur in specific energy levels around an

atom’s nucleus.

• Within each energy level, an electron may move within a particular shape of orbital. An orbital defines

the space in which an electron is most likely to be found. There are different orbital shapes, including

s, p, d and f orbitals.

• Energy diagrams such as Aufbau diagrams are used to show the electron configuration of atoms.

• The electrons in the outermost energy level are called valence electrons.

• The electrons that are not valence electrons are called core electrons.

• Atoms whose outermost energy level is full, are less chemically reactive and therefore more stable,

than those atoms whose outer energy level is not full.

(Presentation: P10030 )

1.37.1 End of chapter exercises

1. Write down only the word/term for each of the following descriptions.

a. The sum of the number of protons and neutrons in an atom

b. The defined space around an atom’s nucleus, where an electron is most likely to be found

2. For each of the following, say whether the statement is True or False. If it is False, re-write the

statement correctly.

a. 2010Ne and 22

10Ne each have 10 protons, 12 electrons and 12 neutrons.

b. The atomic mass of any atom of a particular element is always the same.

c. It is safer to use helium gas rather than hydrogen gas in balloons.

d. Group 1 elements readily form negative ions.

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1.37. SUMMARY CHAPTER 1. THE ATOM

3. Multiple choice questions: In each of the following, choose the one correct answer.

a. The three basic components of an atom are:

a. protons, neutrons, and ions

b. protons, neutrons, and electrons

c. protons, neutrinos, and ions

d. protium, deuterium, and tritium

b. The charge of an atom is...

a. positive

b. neutral

c. negative

c. If Rutherford had used neutrons instead of alpha particles in his scattering experiment, the neu-

trons would...

a. not deflect because they have no charge

b. have deflected more often

c. have been attracted to the nucleus easily

d. have given the same results

d. Consider the isotope 23492 U. Which of the following statements is true?

a. The element is an isotope of 23494 Pu

b. The element contains 234 neutrons

c. The element has the same electron configuration as 23892 U

d. The element has an atomic mass number of 92

e. The electron configuration of an atom of chlorine can be represented using the following notation:

a. 1s22s83s7

b. 1s22s22p63s23p5

c. 1s22s22p63s23p6

d. 1s22s22p5

4. Give the standard notation for the following elements:

a. beryllium

b. carbon-12

c. titanium-48

d. fluorine

5. Give the electron configurations and aufbau diagrams for the following elements:

a. aluminium

b. phosphorus

c. carbon

6. Use standard notation to represent the following elements:

a. argon

b. calcium

c. silver-107

d. bromine-79

7. For each of the following elements give the number of protons, neutrons and electrons in the element:

a. 19578 Pt

b. 4018Ar

c. 5927Co

d. 73Li

e. 115 B

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CHAPTER 1. THE ATOM 1.37. SUMMARY

8. For each of the following elements give the element or number represented by ’x’:

a. 10345 X

b. 35x Cl

c. x4Be

9. Which of the following are isotopes of 2412Mg:

a. 1225Mg

b. 2612Mg

c. 2413Al

10. If a sample contains 69% of copper-63 and 31% of copper-65, calculate the relative atomic mass of

an atom in that sample.

11. Complete the following table:

Element Electron configuration Core electrons Valence electrons

Boron (B)

Calcium (Ca)

Silicon (Si)

Lithium (Li)

Neon (Ne)

Table 4.8

12. Draw aufbau diagrams for the following elements:

a. beryllium

b. sulphur

c. argon

Find the answers with the shortcodes:

(1.) lif (2.) liG (3.a) li7 (3.b) liA (3.c) lio (3.d) lis

(3.e) liH (4.) lg7 (5.) lgG (6.) l44 (7.) l42 (8.) l4T

(9.) l4b (10.) l4j (11.) l4D (12.) lgW

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The Periodic Table

1.38 The arrangement of atoms in the periodic table

(section shortcode: C10031 )

The periodic table of the elements is a method of showing the chemical elements in a table. The elements

are arranged in order of increasing atomic number. Most of the work that was done to arrive at the periodic

table that we know, can be attributed to a man called Dmitri Mendeleev in 1869. Mendeleev was a

Russian chemist who designed the table in such a way that recurring ("periodic") trends in the properties of

the elements could be shown. Using the trends he observed, he even left gaps for those elements that he

thought were ’missing’. He even predicted the properties that he thought the missing elements would have

when they were discovered. Many of these elements were indeed discovered and Mendeleev’s predictions

were proved to be correct.

To show the recurring properties that he had observed, Mendeleev began new rows in his table so that

elements with similar properties were in the same vertical columns, called groups. Each row was referred

to as a period. One important feature to note in the periodic table is that all the non-metals are to the right

of the zig-zag line drawn under the element boron. The rest of the elements are metals, with the exception

of hydrogen which occurs in the first block of the table despite being a non-metal.

Figure 5.1: A simplified diagram showing part of the Periodic Table

You can view the periodic table online2 . The full periodic table is also reproduced at the front of this book.

2http://periodictable.com/

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1.38. THE ARRANGEMENT OF ATOMS IN THE PERIODIC TABLE CHAPTER 1. THE PERIODIC TABLE

1.38.1 Activity: Inventing the periodic table

You are the official chemist for the planet Zog. You have discovered all the same elements that we have

here on Earth, but you don’t have a periodic table. The citizens of Zog want to know how all these elements

relate to each other. How would you invent the periodic table? Think about how you would organize the

data that you have and what properties you would include. Do not simply copy Mendeleev’s ideas, be

creative and come up with some of your own. Research other forms of the periodic table and make one

that makes sense to you. Present your ideas to your class.

1.38.2 Groups in the periodic table

A group is a vertical column in the periodic table and is considered to be the most important way of clas-

sifying the elements. If you look at a periodic table, you will see the groups numbered at the top of each

column. The groups are numbered from left to right starting with 1 and ending with 18. This is the con-

vention that we will use in this book. On some periodic tables you may see that the groups are numbered

from left to right as follows: 1, 2, then an open space which contains the transition elements, followed by

groups 3 to 8. Another way to label the groups is using Roman numerals. In some groups, the elements

display very similar chemical properties and the groups are even given separate names to identify them.

The characteristics of each group are mostly determined by the electron configuration of the atoms of the

element.

• Group 1: These elements are known as the alkali metals and they are very reactive.

Figure 5.2: Electron diagrams for some of the Group 1 elements, with sodium and potasium incomplete; to be

completed as an excersise.

• Group 2: These elements are known as the alkali earth metals. Each element only has two valence

electrons and so in chemical reactions, the group 2 elements tend to lose these electrons so that the

energy shells are complete. These elements are less reactive than those in group 1 because it is

more difficult to lose two electrons than it is to lose one.

• Group 13 elements have three valence electrons.

• Group 16: These elements are sometimes known as the chalcogens. These elements are fairly

reactive and tend to gain electrons to fill their outer shell.

• Group 17: These elements are known as the halogens. Each element is missing just one electron

from its outer energy shell. These elements tend to gain electrons to fill this shell, rather than losing

them. These elements are also very reactive.

• Group 18: These elements are the noble gases. All of the energy shells of the halogens are full and

so these elements are very unreactive.

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Figure 5.3: Electron diagrams for two of the noble gases, helium (He) and neon (Ne).

• Transition metals: The differences between groups in the transition metals are not usually dramatic.

TIP: The number of valence electrons of an element corresponds to its group number

on the periodic table.

NOTE: Group 15 on the periodic table is sometimes called the pnictogens.

Investigation : The properties of elements

Refer to Figure 5.2.

1. Use a periodic table to help you to complete the last two diagrams for sodium (Na) and potassium

(K).

2. What do you notice about the number of electrons in the valence energy level in each case?

3. Explain why elements from group 1 are more reactive than elements from group 2 on the periodic

table (Hint: Think about the ’ionisation energy’).

It is worth noting that in each of the groups described above, the atomic diameter of the elements in-

creases as you move down the group. This is because, while the number of valence electrons is the same

in each element, the number of core electrons increases as one moves down the group.

Khan academy video on the periodic table - 1 (Video: P10032 )

1.38.3 Periods in the periodic table

A period is a horizontal row in the periodic table of the elements. Some of the trends that can be observed

within a period are highlighted below:

• As you move from one group to the next within a period, the number of valence electrons increases

by one each time.

• Within a single period, all the valence electrons occur in the same energy shell. If the period increases,

so does the energy shell in which the valence electrons occur.

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1.38. THE ARRANGEMENT OF ATOMS IN THE PERIODIC TABLE CHAPTER 1. THE PERIODIC TABLE

• In general, the diameter of atoms decreases as one moves from left to right across a period. Consider

the attractive force between the positively charged nucleus and the negatively charged electrons in an

atom. As you move across a period, the number of protons in each atom increases. The number of

electrons also increases, but these electrons will still be in the same energy shell. As the number of

protons increases, the force of attraction between the nucleus and the electrons will increase and the

atomic diameter will decrease.

• Ionisation energy increases as one moves from left to right across a period. As the valence electron

shell moves closer to being full, it becomes more difficult to remove electrons. The opposite is true

when you move down a group in the table because more energy shells are being added. The elec-

trons that are closer to the nucleus ’shield’ the outer electrons from the attractive force of the positive

nucleus. Because these electrons are not being held to the nucleus as strongly, it is easier for them

to be removed and the ionisation energy decreases.

• In general, the reactivity of the elements decreases from left to right across a period.

• The formation of halides follows the general pattern: XCln (where X is any element in a specific group

and n is the number of that specific group.). For example, the formula for the halides of group 1 will

be XCl, for the second group the halides have the formula XCl2 and in the third group the halides

have the formula XCl3. This should be easy to see if you remember the valency of the group and of

the halides.

The formation of oxides show a trend as you move across a period. This should be easy to see if you

think about valency. In the first group all the elements lose an electron to form a cation. So the formula for

an oxide will be X2O. In the second group (moving from left to right across a period) the oxides have the

formula XO. In the third group the oxides have the formula X2O3.

Several other trends may be observed across a period such as density, melting points and boiling points.

These trends are not as obvious to see as the above trends and often show variations to the general trend.

Electron affinity and electronegativity also show some general trends across periods. Electron affinity can

be thought of as how much an element wants electrons. Electron affinity generally increases from left

to right across a period. Electronegativity is the tendency of atoms to attract electrons. The higher the

electronegativity, the greater the atom attracts electrons. Electronegativity generally increases across a

period (from left to right). Electronegativity and electron affinity will be covered in more detail in a later

grade.

You may see periodic tables labeled with s-block, p-block, d-block and f-block. This is simply another way

to group the elements. When we group elements like this we are simply noting which orbitals are being

filled in each block. This method of grouping is not very useful to the work covered at this level.

Using the properties of the groups and the trends that we observe in certain properties (ionization energy,

formation of halides and oxides, melting and boiling points, atomic diameter) we can predict the the prop-

erties of unknown elements. For example, the properties of the unfamiliar elements Francium (Fr), Barium

(Ba), Astatine (At), and Xenon (Xe) can be predicted by knowing their position on the periodic table. Using

the periodic table we can say: Francium (Group 1) is an alkali metal, very reactive and needs to lose 1

electron to obtain a full outer energy shell; Barium (Group 2) is an alkali earth metal and needs to lose 2

electrons to achieve stability; Astatine (Group 7) is a halogen, very reactive and needs to gain 1 electron

to obtain a full outer energy shell; and Xenon (Group 8) is a noble gas and thus stable due to its full outer

energy shell. This is how scientists are able to say what sort of properties the atoms in the last period

have. Almost all of the elements in this period do not occur naturally on earth and are made in laboratories.

These atoms do not exist for very long (they are very unstable and break apart easily) and so measuring

their properties is difficult.

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CHAPTER 1. THE PERIODIC TABLE 1.39. IONISATION ENERGY AND THE PERIODIC TABLE

Exercise: Elements in the periodic table

Refer to the elements listed below:

• Lithium (Li)• Chlorine (Cl)• Magnesium (Mg)

• Neon (Ne)• Oxygen (O)

• Calcium (Ca)

• Carbon (C)

Which of the elements listed above:

1. belongs to Group 1

2. is a halogen

3. is a noble gas

4. is an alkali metal

5. has an atomic number of 12

6. has 4 neutrons in the nucleus of its atoms

7. contains electrons in the 4th energy level

8. has only one valence electron

9. has all its energy orbitals full

10. will have chemical properties that are most similar

11. will form positive ions

Find the answers with the shortcodes:

(1.) liw

1.39 Ionisation Energy and the Periodic Table

(section shortcode: C10033 )

1.39.1 Ions

In the previous section, we focused our attention on the electron configuration of neutral atoms. In a neutral

atom, the number of protons is the same as the number of electrons. But what happens if an atom gains

or loses electrons? Does it mean that the atom will still be part of the same element?

A change in the number of electrons of an atom does not change the type of atom that it is. However,

the charge of the atom will change. If electrons are added, then the atom will become more negative. If

electrons are taken away, then the atom will become more positive. The atom that is formed in either of

these cases is called an ion. Put simply, an ion is a charged atom.

Definition: Ion

An ion is a charged atom. A positively charged ion is called a cation e.g. Na+, and a

negatively charged ion is called an anion e.g. F−. The charge on an ion depends on the

number of electrons that have been lost or gained.

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But how do we know how many electrons an atom will gain or lose? Remember what we said about

stability? We said that all atoms are trying to get a full outer shell. For the elements on the left hand side

of the periodic table the easiest way to do this is to lose electrons and for the elements on the right of the

periodic table the easiest way to do this is to gain electrons. So the elements on the left of the periodic

table will form cations and the elements on the right hand side of the periodic table will form anions. By

doing this the elements can be in the most stable electronic configuration and so be as stable as the noble

gases.

Look at the following examples. Notice the number of valence electrons in the neutral atom, the number of

electrons that are lost or gained and the final charge of the ion that is formed.

Lithium: A lithium atom loses one electron to form a positive ion:

Figure 5.5: The arrangement of electrons in a lithium ion.

In this example, the lithium atom loses an electron to form the cation Li+.

Fluorine: A fluorine atom gains one electron to form a negative ion:

Figure 5.6: The arrangement of electrons in a fluorine ion.

You should have noticed in both these examples that each element lost or gained electrons to make a full

outer shell.

Investigation : The formation of ions

1. Use the diagram for lithium as a guide and draw similar diagrams to show how each of the following

ions is formed:

a. Mg2+

b. Na+

c. Cl−

d. O2+

2. Do you notice anything interesting about the charge on each of these ions? Hint: Look at the number

of valence electrons in the neutral atom and the charge on the final ion.

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CHAPTER 1. THE PERIODIC TABLE 1.39. IONISATION ENERGY AND THE PERIODIC TABLE

Observations:

Once you have completed the activity, you should notice that:

• In each case the number of electrons that is either gained or lost, is the same as the number of

electrons that are needed for the atoms to achieve a full outer energy level.

• If you look at an energy level diagram for sodium (Na), you will see that in a neutral atom, there is

only one valence electron. In order to achieve a full outer energy level, and therefore a more stable

state for the atom, this electron will be lost.

• In the case of oxygen (O), there are six valence electrons. To achieve a full energy level, it makes

more sense for this atom to gain two electrons. A negative ion is formed.

Exercise: The formation of ions

Match the information in column A with the information in column B by writing only the letter (A to I) next to

the question number (1 to 7)

1. A positive ion that has 3 less electrons than its neutral atom A. Mg2+

2. An ion that has 1 more electron than its neutral atom B. Cl−

3. The anion that is formed when bromine gains an electron C. CO2−3

4. The cation that is formed from a magnesium atom D. Al3+

5. An example of a compound ion E. Br2−

6. A positive ion with the electron configuration of argon F. K+

7. A negative ion with the electron configuration of neon G. Mg+

H. O2−

I. Br−

Table 5.1

Find the answers with the shortcodes:

(1.) lid

1.39.2 Ionisation Energy

Ionisation energy is the energy that is needed to remove one electron from an atom. The ionisation energy

will be different for different atoms.

The second ionisation energy is the energy that is needed to remove a second electron from an atom, and

so on. As an energy level becomes more full, it becomes more and more difficult to remove an electron

and the ionisation energy increases. On the Periodic Table of the Elements, a group is a vertical column of

the elements, and a period is a horizontal row. In the periodic table, ionisation energy increases across a

period, but decreases as you move down a group. The lower the ionisation energy, the more reactive the

element will be because there is a greater chance of electrons being involved in chemical reactions. We

will look at this in more detail in the next section.

Trends in ionisation energy

Refer to the data table below which gives the ionisation energy (in kJ ·mol−1) and atomic number (Z) for a

number of elements in the periodic table:

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1.39. IONISATION ENERGY AND THE PERIODIC TABLE CHAPTER 1. THE PERIODIC TABLE

Z Ionisation energy Z Ionisation energy

1 1310 10 2072

2 2360 11 494

3 517 12 734

4 895 13 575

5 797 14 783

6 1087 15 1051

7 1397 16 994

8 1307 17 1250

9 1673 18 1540

Table 5.2

1. Draw a line graph to show the relationship between atomic number (on the x-axis) and ionisation

energy (y-axis).

2. Describe any trends that you observe.

3. Explain why...

a. the ionisation energy for Z = 2 is higher than for Z = 1b. the ionisation energy for Z = 3 is lower than for Z = 2c. the ionisation energy increases between Z = 5 and Z = 7

Find the answers with the shortcodes:

(1.) liv

Khan academy video on periodic table - 2 (Video: P10034 )

By now you should have an appreciation of what the periodic table can tell us. The periodic table does

not just list the elements, but tells chemists what the properties of elements are, how the elements will

combine and many other useful facts. The periodic table is truly an amazing resource. Into one simple

table, chemists have packed so many facts and data that can easily be seen with a glance. The periodic

table is a crucial part of chemistry and you should never go to science class without it. The following

presentation provides a summary of the periodic table

(Presentation: P10035 )

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CHAPTER 1. THE PERIODIC TABLE 1.40. SUMMARY

1.40 Summary

(section shortcode: C10036 )

• Elements are arranged in periods and groups on the periodic table. The elements are arranged

according to increasing atomic number.

• A group is a column on the periodic table containing elements with similar properties. A period is a

row on the periodic table.

• The groups on the periodic table are labeled from 1 to 8. The first group is known as the alkali metals,

the second group is known as the alkali earth metals, the seventh group is known as the halogens

and the eighth group is known as the noble gases. Each group has the same properties.

• Several trends such as ionisation energy and atomic diameter can be seen across the periods of the

periodic table

• An ion is a charged atom. A cation is a positively charged ion and an anion is a negatively charged

ion.

• When forming an ion, an atom will lose or gain the number of electrons that will make its valence

energy level full.

• An element’s ionisation energy is the energy that is needed to remove one electron from an atom.

• Ionisation energy increases across a period in the periodic table.

• Ionisation energy decreases down a group in the periodic table.

1.41 End of chapter exercises

(section shortcode: C10037 )

1. For the following questions state whether they are true or false. If they are false, correct the statement.

a. The group 1 elements are sometimes known as the alkali earth metals.

b. The group 2 elements tend to lose 2 electrons to form cations.

c. The group 8 elements are known as the noble gases.

d. Group 7 elements are very unreactive.

e. The transition elements are found between groups 3 and 4.

2. Give one word or term for each of the following:

a. A positive ion

b. The energy that is needed to remove one electron from an atom

c. A horizontal row on the periodic table

d. A very reactive group of elements that is missing just one electron from their outer shells.

3. For each of the following elements give the ion that will be formed:

a. sodium

b. bromine

c. magnesium

d. oxygen

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1.41. END OF CHAPTER EXERCISES CHAPTER 1. THE PERIODIC TABLE

4. The following table shows the first ionisation energies for the elements of period 1 and 2.

Period Element First ionisation energy (kJ.mol−1)

1 H 1312

He 2372

Li 520

Be 899

B 801

C 1086

2 N 1402

O 1314

F 1681

Ne 2081

Table 5.3

a. What is the meaning of the term first ionisation energy?

b. Identify the pattern of first ionisation energies in a period.

c. Which TWO elements exert the strongest attractive forces on their electrons? Use the data in the

table to give a reason for your answer.

d. Draw Aufbau diagrams for the TWO elements you listed in the previous question and explain why

these elements are so stable.

e. It is safer to use helium gas than hydrogen gas in balloons. Which property of helium makes it a

safer option?

f. ’Group 1 elements readily form positive ions’. Is this statement correct? Explain your answer by

referring to the table.

Find the answers with the shortcodes:

(1.) l4Z (2.) l4K (3.) l4B (4.) li6

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Chemical Bonding

1.42 Chemical Bonding

(section shortcode: C10038 )

When you look at the matter, or physical substances, around you, you will realise that atoms seldom exist

on their own. More often, the things around us are made up of different atoms that have been joined

together. This is called chemical bonding. Chemical bonding is one of the most important processes in

chemistry because it allows all sorts of different molecules and combinations of atoms to form, which then

make up the objects in the complex world around us.

1.43 What happens when atoms bond?

(section shortcode: C10039 )

A chemical bond is formed when atoms are held together by attractive forces. This attraction occurs

when electrons are shared between atoms, or when electrons are exchanged between the atoms that are

involved in the bond. The sharing or exchange of electrons takes place so that the outer energy levels of

the atoms involved are filled and the atoms are more stable. If an electron is shared, it means that it will

spend its time moving in the electron orbitals around both atoms. If an electron is exchanged it means that

it is transferred from one atom to another, in other words one atom gains an electron while the other loses

an electron.

Definition: Chemical bond

A chemical bond is the physical process that causes atoms and molecules to be attracted

to each other, and held together in more stable chemical compounds.

The type of bond that is formed depends on the elements that are involved. In this chapter, we will be

looking at three types of chemical bonding: covalent, ionic and metallic bonding.

You need to remember that it is the valence electrons that are involved in bonding and that atoms will try to

fill their outer energy levels so that they are more stable (or are more like the noble gases which are very

stable).

1.44 Covalent Bonding

(section shortcode: C10040 )

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1.44. COVALENT BONDING CHAPTER 1. CHEMICAL BONDING

1.44.1 The nature of the covalent bond

Covalent bonding occurs between the atoms of non-metals. The outermost orbitals of the atoms overlap

so that unpaired electrons in each of the bonding atoms can be shared. By overlapping orbitals, the outer

energy shells of all the bonding atoms are filled. The shared electrons move in the orbitals around both

atoms. As they move, there is an attraction between these negatively charged electrons and the positively

charged nuclei, and this force holds the atoms together in a covalent bond.

Definition: Covalent bond

Covalent bonding is a form of chemical bonding where pairs of electrons are shared be-

tween atoms.

Below are a few examples. Remember that it is only the valence electrons that are involved in bonding,

and so when diagrams are drawn to show what is happening during bonding, it is only these electrons that

are shown. Circles and crosses are used to represent electrons in different atoms.

Exercise 6.1: Covalent bonding How do hydrogen and chlorine atoms

bond covalently in a molecule of hydrogen chloride?

Solution to Exercise

Step 1. A chlorine atom has 17 electrons, and an electron configura-

tion of 1s22s22p63s23p5. A hydrogen atom has only 1 elec-

tron, and an electron configuration of 1s1.

Step 2. Chlorine has 7 valence electrons. One of these electrons is

unpaired. Hydrogen has 1 valence electron and it is unpaired.

Step 3. The hydrogen atom needs one more electron to complete its

valence shell. The chlorine atom also needs one more elec-

tron to complete its shell. Therefore one pair of electrons

must be shared between the two atoms. In other words, one

electron from the chlorine atom will spend some of its time

orbiting the hydrogen atom so that hydrogen’s valence shell

is full. The hydrogen electron will spend some of its time or-

biting the chlorine atom so that chlorine’s valence shell is also

full. A molecule of hydrogen chloride is formed (Figure 6.1).

Notice the shared electron pair in the overlapping orbitals.

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CHAPTER 1. CHEMICAL BONDING 1.44. COVALENT BONDING

Figure 6.1: Covalent bonding in a molecule of hydrogen chloride

Exercise 6.2: Covalent bonding involving multiple bonds How do ni-

trogen and hydrogen atoms bond to form a molecule of ammonia (NH3)?

Solution to Exercise

Step 1. A nitrogen atom has 7 electrons, and an electron configuration of

1s22s22p3. A hydrogen atom has only 1 electron, and an electron

configuration of 1s1.

Step 2. Nitrogen has 5 valence electrons meaning that 3 electrons are un-

paired. Hydrogen has 1 valence electron and it is unpaired.

Step 3. Each hydrogen atom needs one more electron to complete its va-

lence energy shell. The nitrogen atom needs three more electrons

to complete its valence energy shell. Therefore three pairs of elec-

trons must be shared between the four atoms involved. The ni-

trogen atom will share three of its electrons so that each of the

hydrogen atoms now have a complete valence shell. Each of the

hydrogen atoms will share its electron with the nitrogen atom to

complete its valence shell (Figure 6.2).

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1.44. COVALENT BONDING CHAPTER 1. CHEMICAL BONDING

Figure 6.2: Covalent bonding in a molecule of ammonia

The above examples all show single covalent bonds, where only one pair of electrons is shared between the

same two atoms. If two pairs of electrons are shared between the same two atoms, this is called a double bond.

A triple bond is formed if three pairs of electrons are shared.

Exercise 6.3: Covalent bonding involving a double bond How do

oxygen atoms bond covalently to form an oxygen molecule?

Solution to Exercise

Step 1. Each oxygen atom has 8 electrons, and their electron configuration

is 1s22s22p4.

Step 2. Each oxygen atom has 6 valence electrons, meaning that each

atom has 2 unpaired electrons.

Step 3. Each oxygen atom needs two more electrons to complete its va-

lence energy shell. Therefore two pairs of electrons must be shared

between the two oxygen atoms so that both valence shells are full.

Notice that the two electron pairs are being shared between the

same two atoms, and so we call this a double bond (Figure 6.3).

Figure 6.3: A double covalent bond in an oxygen molecule

You will have noticed in the above examples that the number of electrons that are involved in bonding varies

between atoms. We say that the valency of the atoms is different.

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CHAPTER 1. CHEMICAL BONDING 1.44. COVALENT BONDING

Definition: Valency

The number of electrons in the outer shell of an atom which are able to be used to form

bonds with other atoms.

In the first example, the valency of both hydrogen and chlorine is one, therefore there is a single covalent bond

between these two atoms. In the second example, nitrogen has a valency of three and hydrogen has a valency

of one. This means that three hydrogen atoms will need to bond with a single nitrogen atom. There are three

single covalent bonds in a molecule of ammonia. In the third example, the valency of oxygen is two. This means

that each oxygen atom will form two bonds with another atom. Since there is only one other atom in a molecule

of O2, a double covalent bond is formed between these two atoms.

TIP: There is a relationship between the valency of an element and its position on the

Periodic Table. For the elements in groups 1 to 4, the valency is the same as the group

number. For elements in groups 5 to 7, the valency is calculated by subtracting the

group number from 8. For example, the valency of fluorine (group 7) is 8−7 = 1, while

the valency of calcium (group 2) is 2. Some elements have more than one possible

valency, so you always need to be careful when you are writing a chemical formula.

Often, if there is more than one possibility in terms of valency, the valency will be

written in a bracket after the element symbol e.g. carbon (IV) oxide, means that in this

molecule carbon has a valency of 4.

Covalent bonding and valency

1. Explain the difference between the valence electrons and the valency of an element.

2. Complete the table below by filling in the number of valence electrons and the valency for each of the

elements shown:

Element No. of valence electrons No. of electrons needed to fill outer shell Valency

F

Ar

C

N

O

Table 6.1

3. Draw simple diagrams to show how electrons are arranged in the following covalent molecules:

a. Water (H2O)

b. Chlorine (Cl2)

Find the answers with the shortcodes:

(1.) lOY (2.) lOr (3.) lO1

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1.45. LEWIS NOTATION AND MOLECULAR STRUCTURE CHAPTER 1. CHEMICAL BONDING

1.44.2 Properties of covalent compounds

Covalent compounds have several properties that distinguish them from ionic compounds and metals. These

properties are:

1. The melting and boiling points of covalent compounds is generally lower than that for ionic compounds.

2. Covalent compounds are generally more flexible than ionic compounds. The molecules in covalent com-

pounds are able to move around to some extent and can sometimes slide over each other (as is the case

with graphite, this is why the lead in your pencil feels slightly slippery). In ionic compounds all the ions are

tightly held in place.

3. Covalent compounds generally are not very soluble in water.

4. Covalent compounds generally do not conduct electricity when dissolved in water. This is because they do

not dissociate as ionic compounds do.

1.45 Lewis notation and molecular structure

(section shortcode: C10041 )

Although we have used diagrams to show the structure of molecules, there are other forms of notation that can

be used, such as Lewis notation and Couper notation. Lewis notation uses dots and crosses to represent the

valence electrons on different atoms. The chemical symbol of the element is used to represent the nucleus and

the core electrons of the atom.

So, for example, a hydrogen atom would be represented like this:

A chlorine atom would look like this:

A molecule of hydrogen chloride would be shown like this:

The dot and cross in between the two atoms, represent the pair of electrons that are shared in the covalent bond.

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CHAPTER 1. CHEMICAL BONDING 1.45. LEWIS NOTATION AND MOLECULAR STRUCTURE

Exercise 6.4: Lewis notation: Simple molecules Represent the

molecule H2O using Lewis notation

Solution to Exercise

Step 1. The electron configuration of hydrogen is 1s1 and the electron con-

figuration for oxygen is 1s22s22p4. Each hydrogen atom has one

valence electron, which is unpaired, and the oxygen atom has six

valence electrons with two unpaired.

Step 2. The water molecule is represented below.

Exercise 6.5: Lewis notation: Molecules with multiple bonds Rep-

resent the molecule HCN (hydrogen cyanide) using Lewis notation

Solution to Exercise

Step 1. The electron configuration of hydrogen is 1s1, the electron config-

uration of nitrogen is 1s22s22p3 and for carbon is 1s22s22p2. This

means that hydrogen has one valence electron which is unpaired,

carbon has four valence electrons, all of which are unpaired, and

nitrogen has five valence electrons, three of which are unpaired.

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1.45. LEWIS NOTATION AND MOLECULAR STRUCTURE CHAPTER 1. CHEMICAL BONDING

Step 2. The HCN molecule is represented below. Notice the three electron

pairs between the nitrogen and carbon atom. Because these three

covalent bonds are between the same two atoms, this is a triple

bond.

Exercise 6.6: Lewis notation: Atoms with variable valencies Rep-

resent the molecule H2S (hydrogen sulphide) using Lewis notation

Solution to Exercise

Step 1. Hydrogen has an electron configuration of 1s1 and sulphur has an

electron configuration of 1s22s22p63s23p4. Each hydrogen atom

has one valence electron which is unpaired, and sulphur has six va-

lence electrons. Although sulphur has a variable valency, we know

that the sulphur will be able to form 2 bonds with the hydrogen

atoms. In this case, the valency of sulphur must be two.

Step 2. The H2S molecule is represented below.

Another way of representing molecules is using Couper notation. In this case, only the electrons that are involved

in the bond between the atoms are shown. A line is used for each covalent bond. Using Couper notation, a

molecule of water and a molecule of HCN would be represented as shown in Figure 6.13 and Figure 6.14 below.

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CHAPTER 1. CHEMICAL BONDING 1.45. LEWIS NOTATION AND MOLECULAR STRUCTURE

Figure 6.13: A water molecule represented using Couper notation

Figure 6.14: A molecule of HCN represented using Couper notation

1.45.1 Atomic bonding and Lewis notation

1. Represent each of the following atoms using Lewis notation:

a. beryllium

b. calcium

c. lithium

2. Represent each of the following molecules using Lewis notation:

a. bromine gas (Br2)

b. carbon dioxide (CO2)

3. Which of the two molecules listed above contains a double bond?

4. Two chemical reactions are described below.

• nitrogen and hydrogen react to form NH3

• carbon and hydrogen bond to form a molecule of CH4

For each reaction, give:

a. the valency of each of the atoms involved in the reaction

b. the Lewis structure of the product that is formed

c. the chemical formula of the product

d. the name of the product

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1.46. IONIC BONDING CHAPTER 1. CHEMICAL BONDING

5. A chemical compound has the following Lewis notation:

a. How many valence electrons does element Y have?

b. What is the valency of element Y?

c. What is the valency of element X?

d. How many covalent bonds are in the molecule?

e. Suggest a name for the elements X and Y.

Find the answers with the shortcodes:

(1.) lOC (2.) lOa (3.) lOa (4.) lOx (5.) lOc

1.46 Ionic Bonding

(section shortcode: C10042 )

1.46.1 The nature of the ionic bond

You will remember that when atoms bond, electrons are either shared or they are transferred between the atoms

that are bonding. In covalent bonding, electrons are shared between the atoms. There is another type of bonding,

where electrons are transferred from one atom to another. This is called ionic bonding.

Ionic bonding takes place when the difference in electronegativity between the two atoms is more than 1,7. This

usually happens when a metal atom bonds with a non-metal atom. When the difference in electronegativity is

large, one atom will attract the shared electron pair much more strongly than the other, causing electrons to be

transferred from one atom to the other.

Definition: Ionic bond

An ionic bond is a type of chemical bond based on the electrostatic forces between two

oppositely-charged ions. When ionic bonds form, a metal donates one or more electrons,

due to having a low electronegativity, to form a positive ion or cation. The non-metal atom

has a high electronegativity, and therefore readily gains electrons to form a negative ion or

anion. The two ions are then attracted to each other by electrostatic forces.

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CHAPTER 1. CHEMICAL BONDING 1.46. IONIC BONDING

Example 1:

In the case of NaCl, the difference in electronegativity is 2,1. Sodium has only one valence electron, while

chlorine has seven. Because the electronegativity of chlorine is higher than the electronegativity of sodium,

chlorine will attract the valence electron of the sodium atom very strongly. This electron from sodium is transferred

to chlorine. Sodium loses an electron and forms an Na+ ion. Chlorine gains an electron and forms an Cl− ion.

The attractive force between the positive and negative ion holds the molecule together.

The balanced equation for the reaction is:

Na + Cl → NaCl (6.1)

This can be represented using Lewis notation:

Figure 6.16: Ionic bonding in sodium chloride

Example 2:

Another example of ionic bonding takes place between magnesium (Mg) and oxygen (O) to form magnesium

oxide (MgO). Magnesium has two valence electrons and an electronegativity of 1,2, while oxygen has six valence

electrons and an electronegativity of 3,5. Since oxygen has a higher electronegativity, it attracts the two valence

electrons from the magnesium atom and these electrons are transferred from the magnesium atom to the oxygen

atom. Magnesium loses two electrons to form Mg2+, and oxygen gains two electrons to form O2−. The attractive

force between the oppositely charged ions is what holds the molecule together.

The balanced equation for the reaction is:

2Mg + O2 → 2MgO (6.2)

Because oxygen is a diatomic molecule, two magnesium atoms will be needed to combine with one oxygen

molecule (which has two oxygen atoms) to produce two molecules of magnesium oxide (MgO).

Figure 6.17: Ionic bonding in magnesium oxide

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1.46. IONIC BONDING CHAPTER 1. CHEMICAL BONDING

TIP: Notice that the number of electrons that is either lost or gained by an atom during

ionic bonding, is the same as the valency of that element

Ionic compounds

1. Explain the difference between a covalent and an ionic bond.

2. Magnesium and chlorine react to form magnesium chloride.

a. What is the difference in electronegativity between these two elements?

b. Give the chemical formula for:

i. a magnesium ion

ii. a chloride ion

iii. the ionic compound that is produced during this reaction

c. Write a balanced chemical equation for the reaction that takes place.

3. Draw Lewis diagrams to represent the following ionic compounds:

a. sodium iodide (NaI)b. calcium bromide (CaBr2)

c. potassium chloride (KCl)

Find the answers with the shortcodes:

(1.) lOq (2.) lOl (3.) lOi

1.46.2 The crystal lattice structure of ionic compounds

Ionic substances are actually a combination of lots of ions bonded together into a giant molecule. The arrange-

ment of ions in a regular, geometric structure is called a crystal lattice. So in fact NaCl does not contain one Naand one Cl ion, but rather a lot of these two ions arranged in a crystal lattice where the ratio of Na to Cl ions is

1:1. The structure of a crystal lattice is shown in Figure 6.18.

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CHAPTER 1. CHEMICAL BONDING 1.47. METALLIC BONDS

Figure 6.18: The crystal lattice arrangement in an ionic compound (e.g. NaCl)

1.46.3 Properties of Ionic Compounds

Ionic compounds have a number of properties:

• Ions are arranged in a lattice structure

• Ionic solids are crystalline at room temperature

• The ionic bond is a strong electrical attraction. This means that ionic compounds are often hard and have

high melting and boiling points

• Ionic compounds are brittle, and bonds are broken along planes when the compound is stressed

• Solid crystals don’t conduct electricity, but ionic solutions do

1.47 Metallic bonds

(section shortcode: C10043 )

1.47.1 The nature of the metallic bond

The structure of a metallic bond is quite different from covalent and ionic bonds. In a metal bond, the valence

electrons are delocalised, meaning that an atom’s electrons do not stay around that one nucleus. In a metallic

bond, the positive atomic nuclei (sometimes called the ’atomic kernels’) are surrounded by a sea of delocalised

electrons which are attracted to the nuclei (Figure 6.19).

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1.47. METALLIC BONDS CHAPTER 1. CHEMICAL BONDING

Definition: Metallic bond

Metallic bonding is the electrostatic attraction between the positively charged atomic nuclei

of metal atoms and the delocalised electrons in the metal.

Figure 6.19: Positive atomic nuclei (+) surrounded by delocalised electrons (•)

1.47.2 The properties of metals

Metals have several unique properties as a result of this arrangement:

• Thermal conductors Metals are good conductors of heat and are therefore used in cooking utensils such

as pots and pans. Because the electrons are loosely bound and are able to move, they can transport heat

energy from one part of the material to another.

• Electrical conductors Metals are good conductors of electricity, and are therefore used in electrical conduct-

ing wires. The loosely bound electrons are able to move easily and to transfer charge from one part of the

material to another.

• Shiny metallic lustre Metals have a characteristic shiny appearance and are often used to make jewellery.

The loosely bound electrons are able to absorb and reflect light at all frequencies, making metals look

polished and shiny.

• Malleable and ductile This means that they can be bent into shape without breaking (malleable) and can be

stretched into thin wires (ductile) such as copper, which can then be used to conduct electricity. Because

the bonds are not fixed in a particular direction, atoms can slide easily over one another, making metals

easy to shape, mould or draw into threads.

• Melting point Metals usually have a high melting point and can therefore be used to make cooking pots and

other equipment that needs to become very hot, without being damaged. The high melting point is due to

the high strength of metallic bonds.

• Density Metals have a high density because their atoms are packed closely together.

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CHAPTER 1. CHEMICAL BONDING 1.47. METALLIC BONDS

1.47.3 Activity: Building models

Using coloured balls and sticks (or any other suitable materials) build models of each type of bonding. Think

about how to represent each kind of bonding. For example, covalent bonding could be represented by simply

connecting the balls with sticks to represent the molecules, while for ionic bonding you may wish to construct part

of the crystal lattice. Do some research on types of crystal lattices (although the section on ionic bonding only

showed the crystal lattice for sodium chloride, many other types of lattices exist) and try to build some of these.

Share your findings with your class and compare notes to see what types of crystal lattices they found. How

would you show metallic bonding?

You should spend some time doing this activity as it will really help you to understand how atoms combine to form

molecules and what the differences are between the types of bonding.

Khan academy video on bonding - 1 (Video: P10044 )

1.47.4 Chemical bonding

1. Give two examples of everyday objects that contain..

a. covalent bonds

b. ionic bonds

c. metallic bonds

2. Complete the table which compares the different types of bonding:

Covalent Ionic Metallic

Types of atoms involved

Nature of bond between atoms

Melting Point (high/low)

Conducts electricity? (yes/no)

Other properties

Table 6.2

3. Complete the table below by identifying the type of bond (covalent, ionic or metallic) in each of the com-

pounds:

Molecular formula Type of bond

H2SO4

FeS

NaI

MgCl2

Zn

Table 6.3

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1.48. WRITING CHEMICAL FORMULAE CHAPTER 1. CHEMICAL BONDING

4. Which of these substances will conduct electricity most effectively? Give a reason for your answer.

5. Use your knowledge of the different types of bonding to explain the following statements:

a. Swimming during an electric storm (i.e. where there is lightning) can be very dangerous.

b. Most jewellery items are made from metals.

c. Plastics are good insulators.

Find the answers with the shortcodes:

(1.) l3h (2.) l3u (3.) l3J (4.) l3J (5.) l3S

1.48 Writing chemical formulae

(section shortcode: C10045 )

1.48.1 The formulae of covalent compounds

To work out the formulae of covalent compounds, we need to use the valency of the atoms in the compound. This

is because the valency tells us how many bonds each atom can form. This in turn can help to work out how many

atoms of each element are in the compound, and therefore what its formula is. The following are some examples

where this information is used to write the chemical formula of a compound.

Exercise 6.7: Formulae of covalent compounds Write the chemical

formula for water

Solution to Exercise

Step 1. A molecule of water contains the elements hydrogen and oxygen.

Step 2. The valency of hydrogen is 1 and the valency of oxygen is 2. This

means that oxygen can form two bonds with other elements and

each of the hydrogen atoms can form one.

Step 3. Using the valencies of hydrogen and oxygen, we know that in a

single water molecule, two hydrogen atoms will combine with one

oxygen atom. The chemical formula for water is therefore:

H2O.

Exercise 6.8: Formulae of covalent compounds Write the chemical

formula for magnesium oxide

Solution to Exercise

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CHAPTER 1. CHEMICAL BONDING 1.48. WRITING CHEMICAL FORMULAE

Step 1. A molecule of magnesium oxide contains the elements magnesium

and oxygen.

Step 2. The valency of magnesium is 2, while the valency of oxygen is also

2. In a molecule of magnesium oxide, one atom of magnesium will

combine with one atom of oxygen.

Step 3. The chemical formula for magnesium oxide is therefore:

MgO

Exercise 6.9: Formulae of covalent compounds Write the chemical

formula for copper (II) chloride.

Solution to Exercise

Step 1. A molecule of copper (II) chloride contains the elements copper

and chlorine.

Step 2. The valency of copper is 2, while the valency of chlorine is 1. In a

molecule of copper (II) chloride, two atoms of chlorine will combine

with one atom of copper.

Step 3. The chemical formula for copper (II) chloride is therefore:

CuCl2

1.48.2 The formulae of ionic compounds

The overall charge of an ionic compound will always be zero and so the negative and positive charge must be

the same size. We can use this information to work out what the chemical formula of an ionic compound is if we

know the charge on the individual ions. In the case of NaCl for example, the charge on the sodium is +1 and the

charge on the chlorine is −1. The charges balance (+1 − 1 = 0) and therefore the ionic compound is neutral.

In MgO, magnesium has a charge of +2 and oxygen has a charge of −2. Again, the charges balance and the

compound is neutral. Positive ions are called cations and negative ions are called anions.

Some ions are made up of groups of atoms, and these are called compound ions. It is a good idea to learn the

compound ions that are shown in Table 6.4

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1.48. WRITING CHEMICAL FORMULAE CHAPTER 1. CHEMICAL BONDING

Name of compound ion formula Name of compound ion formula

Carbonate CO2−3 Nitrate NO−

2

Sulphate SO2−4 Hydrogen sulphite HSO−

3

Hydroxide OH− Hydrogen sulphate HSO−4

Ammonium NH+4 Dihydrogen phosphate H2PO

−4

Nitrate NO−3 Hypochlorite ClO−

Hydrogen carbonate HCO−3 Acetate (ethanoate) CH3COO−

Phosphate PO3−4 Oxalate C2O

2−4

Chlorate ClO−3 Oxide O2−

Cyanide CN− Peroxide O2−2

Chromate CrO2−4 Sulphide S2−

Permanganate MnO−4 Sulphite SO2−

3

Thiosulphate S2O2−3 Manganate MnO2−

4

Phosphide P3− Hydrogen phosphate HPO3−4

Table 6.4: Table showing common compound ions and their formulae

In the case of ionic compounds, the valency of an ion is the same as its charge (Note: valency is always expressed

as a positive number e.g. valency of the chloride ion is 1 and not -1). Since an ionic compound is always neutral,

the positive charges in the compound must balance out the negative. The following are some examples:

Exercise 6.10: Formulae of ionic compounds Write the chemical

formula for potassium iodide.

Solution to Exercise

Step 1. Potassium iodide contains potassium and iodide ions.

Step 2. Potassium iodide contains the ions K+ (valency = 1; charge = +1)

and I− (valency = 1; charge = −1). In order to balance the charge

in a single molecule, one atom of potassium will be needed for every

one atom of iodine.

Step 3. The chemical formula for potassium iodide is therefore:

KI

Exercise 6.11: Formulae of ionic compounds Write the chemical

formula for sodium sulphate.

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CHAPTER 1. CHEMICAL BONDING 1.48. WRITING CHEMICAL FORMULAE

Solution to Exercise

Step 1. Sodium sulphate contains sodium ions and sulphate ions.

Step 2. Na+ (valency = 1; charge = +1) and SO2−4 (valency = 2; charge =

−2).

Step 3. Two sodium ions will be needed to balance the charge of the sul-

phate ion. The chemical formula for sodium sulphate is therefore:

Na2SO4

Exercise 6.12: Formulae of ionic compounds Write the chemical

formula for calcium hydroxide.

Solution to Exercise

Step 1. Calcium hydroxide contains calcium ions and hydroxide ions.

Step 2. Calcium hydroxide contains the ions Ca2+ (charge = +2) and OH−

(charge = −1). In order to balance the charge in a single molecule,

two hydroxide ions will be needed for every ion of calcium.

Step 3. The chemical formula for calcium hydroxide is therefore:

Ca (OH)2

NOTE: Notice how in the last example we wrote OH− inside brackets. We do this to

indicate that OH− is a complex ion and that there are two of these ions bonded to one

calcium ion.

Chemical formulae

1. Copy and complete the table below:

Compound Cation Anion Formula

Na+ Cl−

potassium bromide Br−

NH+4 Cl−

potassium chromate

PbI

potassium permanganate

calcium phosphate

Table 6.5

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1.49. CHEMICAL COMPOUNDS: NAMES AND MASSES CHAPTER 1. CHEMICAL BONDING

2. Write the chemical formula for each of the following compounds:

a. hydrogen cyanide

b. carbon dioxide

c. sodium carbonate

d. ammonium hydroxide

e. barium sulphate

Find the answers with the shortcodes:

(1.) l3t (2.) l3z

1.49 Chemical compounds: names and masses

(section shortcode: C10046 )

In Giving names and formulae to substances (Section 1.4: Giving names and formulae to substances) the names

of chemical compounds was revised. The relative molecular mass for covalent molecules is simply the sum of

the relative atomic masses of each of the individual atoms in that compound. For ionic compounds we use the

formula of the compound to work out a relative formula mass. We ignore the fact that there are many molecules

linked together to form a crystal lattice. For example NaCl has a relative formula mass of 58 g ·mol−1 .

(Presentation: P10047 )

1.50 Summary

(section shortcode: C10048 )

• A chemical bond is the physical process that causes atoms and molecules to be attracted together and to

be bound in new compounds.

• Atoms are more reactive, and therefore more likely to bond, when their outer electron orbitals are not full.

Atoms are less reactive when these outer orbitals contain the maximum number of electrons. This explains

why the noble gases do not combine to form molecules.

• When atoms bond, electrons are either shared or exchanged.

• Covalent bonding occurs between the atoms of non-metals and involves a sharing of electrons so that the

orbitals of the outermost energy levels in the atoms are filled.

• The valency of an atom is the number of electrons in the outer shell of that atom and valence electrons are

able to form bonds with other atoms.

• A double or triple bond occurs if there are two or three electron pairs that are shared between the same

two atoms.

• A dative covalent bond is a bond between two atoms in which both the electrons that are shared in the

bond come from the same atom.

• Lewis and Couper notation are two ways of representing molecular structure. In Lewis notation, dots and

crosses are used to represent the valence electrons around the central atom. In Couper notation, lines are

used to represent the bonds between atoms.

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CHAPTER 1. CHEMICAL BONDING 1.50. SUMMARY

• An ionic bond occurs between atoms where the difference in electronegativity is greater than 1,7. An

exchange of electrons takes place and the atoms are held together by the electrostatic force of attraction

between oppositely-charged ions.

• Ionic solids are arranged in a crystal lattice structure.

• Ionic compounds have a number of specific properties, including their high melting and boiling points, brittle

nature, the lattice structure of solids and the ability of ionic solutions to conduct electricity.

• A metallic bond is the electrostatic attraction between the positively charged nuclei of metal atoms and the

delocalised electrons in the metal.

• Metals also have a number of properties, including their ability to conduct heat and electricity, their metallic

lustre, the fact that they are both malleable and ductile, and their high melting point and density.

• The valency of atoms, and the way they bond, can be used to determine the chemical formulae of com-

pounds.

1.50.1 End of chapter exercises

1. Explain the meaning of each of the following terms

a. Valency

b. Covalent bond

2. Which ONE of the following best describes the bond formed between an H+ ion and the NH3 molecule?

a. Covalent bond

b. Dative covalent (coordinate covalent) bond

c. Ionic Bond

d. Hydrogen Bond

3. Which of the following reactions will not take place? Explain your answer.

a. H+H → H2

b. Ne + Ne → Ne2c. Cl + Cl → Cl2

4. Draw the Lewis structure for each of the following:

a. calcium

b. iodine (Hint: Which group is it in? It will be similar to others in that group)

c. hydrogen bromide (HBr)d. nitrogen dioxide (NO2)

5. Given the following Lewis structure, where X and Y each represent a different element...

a. What is the valency of X?

b. What is the valency of Y?

c. Which elements could X and Y represent?

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1.50. SUMMARY CHAPTER 1. CHEMICAL BONDING

6. A molecule of ethane has the formula C2H6. Which of the following diagrams (Couper notation) accurately

represents this molecule?

7. Potassium dichromate is dissolved in water.

a. Give the name and chemical formula for each of the ions in solution.

b. What is the chemical formula for potassium dichromate?

Find the answers with the shortcodes:

(1.) lgV (2.) l37 (3.) l36 (4.) l3H (5.) l3s (6.) l3o

(7.) l3A

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What are the objects around us made of

1.51 Introduction: The atom as the building block of matter

(section shortcode: C10049 )

We have now seen that different materials have different properties. Some materials are metals and some are

non-metals; some are electrical or thermal conductors, while others are not. Depending on the properties of these

materials, they can be used in lots of useful applications. But what is it exactly that makes up these materials?

In other words, if we were to break down a material into the parts that make it up, what would we find? And how

is it that a material’s microscopic structure (the small parts that make up the material) is able to give it all these

different properties?

The answer lies in the smallest building block of matter: the atom. It is the type of atoms, and the way in which

they are arranged in a material, that affects the properties of that substance. This is similar to building materials.

We can use bricks, steel, cement, wood, straw (thatch), mud and many other things to build structures from. In

the same way that the choice of building material affects the properties of the structure, so the atoms that make

up matter affect the properties of matter.

It is not often that substances are found in atomic form (just as you seldom find a building or structure made from

one building material). Normally, atoms are bonded (joined) to other atoms to form compounds or molecules.

It is only in the noble gases (e.g. helium, neon and argon) that atoms are found individually and are not bonded

to other atoms. We looked at some of the reasons for this in earlier chapters.

1.52 Compounds

(section shortcode: C10050 )

Definition: Compound

A compound is a group of two or more atoms that are attracted to each other by relatively

strong forces or bonds.

Almost everything around us is made up of molecules. The only substances that are not made of molecules,

but instead are individual atoms are the noble gases. Water is made up of molecules, each of which has two

hydrogen atoms joined to one oxygen atom. Oxygen is a molecule that is made up of two oxygen atoms that are

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1.52. COMPOUNDS CHAPTER 1. WHAT ARE THE OBJECTS AROUND US MADE OF

joined to one another. Even the food that we eat is made up of molecules that contain atoms of elements such

as carbon, hydrogen and oxygen that are joined to one another in different ways. All of these are known as small

molecules because there are only a few atoms in each molecule. Giant molecules are those where there may

be millions of atoms per molecule. Examples of giant molecules are diamonds, which are made up of millions of

carbon atoms bonded to each other and metals, which are made up of millions of metal atoms bonded to each

other.

As we learnt in Chapter 5 atoms can share electrons to form covalent bonds or exchange electrons to form ionic

bonds. Covalently bonded substances are known as molecular compounds. Ionically bonded substances are

known as ionic compounds. We also learnt about metallic bonding. In a metal the atoms lose their outermost

electrons to form positively charged ions that are arranged in a lattice, while the outermost electrons are free to

move amongst the spaces of the lattice.

We can classify covalent molecules into covalent molecular structures and covalent network structures. Covalent

molecular structures are simply individual covalent molecules and include water, oxygen, sulphur (S8) and buck-

minsterfullerene (C60). All covalent molecular structures are simple molecules. Covalent network structures are

giant lattices of covalently bonded molecules, similar to the ionic lattice. Examples include diamond, graphite and

silica (SiO2). All covalent network structures are giant molecules.

Examples of ionic substances are sodium chloride (NaCl) and potassium permanganate (KMnO4). Examples

of metals are copper, zinc, titanium, gold, etc.

1.52.1 Representing molecules

The structure of a molecule can be shown in many different ways. Sometimes it is easiest to show what a

molecule looks like by using different types of diagrams, but at other times, we may decide to simply represent a

molecule using its chemical formula or its written name.

1. Using formulae to show the structure of a molecule. A chemical formula is an abbreviated (shortened)

way of describing a molecule, or some other chemical substance. In the chapter on classification of matter,

we saw how chemical compounds can be represented using element symbols from the Periodic Table. A

chemical formula can also tell us the number of atoms of each element that are in a molecule and their ratio

in that molecule. For example, the chemical formula for a molecule of carbon dioxide is CO2 The formula

above is called the molecular formula of that compound. The formula tells us that in one molecule of

carbon dioxide, there is one atom of carbon and two atoms of oxygen. The ratio of carbon atoms to oxygen

atoms is 1:2.

Definition: Molecular formula

This is a concise way of expressing information about the atoms that make up a particular

chemical compound. The molecular formula gives the exact number of each type of atom

in the molecule.

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CHAPTER 1. WHAT ARE THE OBJECTS AROUND US MADE OF 1.52. COMPOUNDS

A molecule of glucose has the molecular formula: C6H12O6. In each glucose molecule, there are six carbon

atoms, twelve hydrogen atoms and six oxygen atoms. The ratio of carbon:hydrogen:oxygen is 6:12:6. We

can simplify this ratio to write 1:2:1, or if we were to use the element symbols, the formula would be written

as CH2O. This is called the empirical formula of the molecule.

Definition: Empirical formula

This is a way of expressing the relative number of each type of atom in a chemical com-

pound. In most cases, the empirical formula does not show the exact number of atoms,

but rather the simplest ratio of the atoms in the compound.

The empirical formula is useful when we want to write the formula for a giant molecule. Since giant molecules

may consist of millions of atoms, it is impossible to say exactly how many atoms are in each molecule. It

makes sense then to represent these molecules using their empirical formula. So, in the case of a metal

such as copper, we would simply write Cu, or if we were to represent a molecule of sodium chloride, we

would simply write NaCl. Chemical formulae therefore tell us something about the types of atoms that are

in a molecule and the ratio in which these atoms occur in the molecule, but they don’t give us any idea of

what the molecule actually looks like, in other words its shape. To show the shape of molecules we can

represent molecules using diagrams. Another type of formula that can be used to describe a molecule is

its structural formula. A structural formula uses a graphical representation to show a molecule’s structure

(Figure 7.1).

Figure 7.1: Diagram showing (a) the molecular, (b) the empirical and (c) the structural formula of isobutane

2. Using diagrams to show the structure of a molecule Diagrams of molecules are very useful because

they help us to picture how the atoms are arranged in the molecule and they help us to see the shape of the

molecule. There are two types of diagrams that are commonly used:

• Ball and stick models This is a 3-dimensional molecular model that uses ’balls’ to represent atoms and

’sticks’ to represent the bonds between them. The centres of the atoms (the balls) are connected by

straight lines which represent the bonds between them. A simplified example is shown in Figure 7.2.

Figure 7.2: A ball and stick model of a water molecule

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1.52. COMPOUNDS CHAPTER 1. WHAT ARE THE OBJECTS AROUND US MADE OF

• Space-filling model This is also a 3-dimensional molecular model. The atoms are represented by

spheres. Figure 7.3 and Figure 7.4 are some examples of simple molecules that are represented in

different ways.

Figure 7.3: A space-filling model and structural formula of a water molecule. Each molecule is made up of two

hydrogen atoms that are attached to one oxygen atom. This is a simple molecule.

Figure 7.4: A space-filling model and structural formula of a molecule of ammonia. Each molecule is made up

of one nitrogen atom and three hydrogen atoms. This is a simple molecule.

Figure 7.5 shows the bonds between the carbon atoms in diamond, which is a giant molecule. Each carbon

atom is joined to four others, and this pattern repeats itself until a complex lattice structure is formed. Each

black ball in the diagram represents a carbon atom, and each line represents the bond between two carbon

atoms. Note that the carbon atoms on the edges are actually bonded to four carbon atoms, but some of

these carbon atoms have been omitted.

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CHAPTER 1. WHAT ARE THE OBJECTS AROUND US MADE OF 1.52. COMPOUNDS

Figure 7.5: Diagrams showing the microscopic structure of diamond. The diagram on the left shows part of a

diamond lattice, made up of numerous carbon atoms. The diagram on the right shows how each carbon atom in

the lattice is joined to four others. This forms the basis of the lattice structure. Diamond is a giant molecule.

NOTE: Diamonds are most often thought of in terms of their use in the jewellery in-

dustry. However, about 80% of mined diamonds are unsuitable for use as gemstones

and are therefore used in industry because of their strength and hardness. These

properties of diamonds are due to the strong covalent bonds (covalent bonding will be

explained later) between the carbon atoms in diamond. The most common uses for

diamonds in industry are in cutting, drilling, grinding, and polishing.

Alteredqualia is a website3 that allows you to view several molecules. You do not need to know these molecules,

this is simply to allow you to see one way of representing molecules.

Atoms and molecules

1. In each of the following, say whether the chemical substance is made up of single atoms, simple molecules

or giant molecules.

a. ammonia gas (NH3)

b. zinc metal (Zn)

c. graphite (C)

d. nitric acid (HNO3)

e. neon gas (He)

3http://alteredqualia.com/canvasmol/

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1.53. SUMMARY CHAPTER 1. WHAT ARE THE OBJECTS AROUND US MADE OF

2. Refer to the diagram below and then answer the questions that follow:

a. Identify the molecule.

b. Write the molecular formula for the molecule.

c. Is the molecule a simple or giant molecule?

3. Represent each of the following molecules using its chemical formula, structural formula and ball and stick

model.

a. Hydrogen

b. Ammonia

c. sulphur dioxide

Find the answers with the shortcodes:

(1.) li5 (2.) liN (3.) liR

1.53 Summary

(section shortcode: C10051 )

• The smallest unit of matter is the atom. Atoms can combine to form molecules.

• A compound is a group of two or more atoms that are attracted to each other by chemical bonds.

• A small molecule consists of a few atoms per molecule. A giant molecule consists of millions of atoms

per molecule, for example metals and diamonds.

• The structure of a molecule can be represented in a number of ways.

• The chemical formula of a molecule is an abbreviated way of showing a molecule, using the symbols for

the elements in the molecule. There are two types of chemical formulae: molecular and empirical formula.

• The molecular formula of a molecule gives the exact number of atoms of each element that are in the

molecule.

• The empirical formula of a molecule gives the relative number of atoms of each element in the molecule.

• Molecules can also be represented using diagrams.

• A ball and stick diagram is a 3-dimensional molecular model that uses ’balls’ to represent atoms and ’sticks’

to represent the bonds between them.

• A space-filling model is also a 3-dimensional molecular model. The atoms are represented by spheres.

• In a molecule, atoms are held together by chemical bonds or intramolecular forces. Covalent bonds,

ionic bonds and metallic bonds are examples of chemical bonds.

• A covalent bond exists between non-metal atoms. An ionic bond exists between non-metal and metal

atoms and a metallic bond exists between metal atoms.

• Intermolecular forces are the bonds that hold molecules together.

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CHAPTER 1. WHAT ARE THE OBJECTS AROUND US MADE OF 1.53. SUMMARY

1.53.1 End of chapter exercises

1. Give one word or term for each of the following descriptions.

a. A composition of two or more atoms that act as a unit.

b. Chemical formula that gives the relative number of atoms of each element that are in a molecule.

2. Give a definition for each of the following terms: descriptions.

a. molecule

b. Ionic compound

c. Covalent network structure

d. Empirical formula

e. Ball-and-stick model

3. Ammonia, an ingredient in household cleaners, can be broken down to form one part nitrogen (N) and three

parts hydrogen (H). This means that ammonia...

a. is a colourless gas

b. is not a compound

c. cannot be an element

d. has the formula N3H

4. Represent each of the following molecules using its chemical formula, its structural formula and the ball-

and-stick model:

a. nitrogen

b. carbon dioxide

c. methane

d. argon

Find the answers with the shortcodes:

(1.) l2e (2.) l2M (3.) liV (4.) l2L

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Physical and Chemical Change

1.54 Physical and Chemical Change - Grade 10

(section shortcode: C10052 )

Matter is all around us. The desks we sit at, the air we breathe and the water we drink are all examples of matter.

But matter doesn’t always stay the same. It can change in many different ways. In this chapter, we are going to

take a closer look at physical and chemical changes that occur in matter.

1.55 Physical changes in matter

(section shortcode: C10053 )

A physical change is one where the particles of the substances that are involved in the change are not broken

up in any way. When water is heated for example, the temperature and energy of the water molecules increases

and the liquid water evaporates to form water vapour. When this happens, some kind of change has taken place,

but the molecular structure of the water has not changed. This is an example of a physical change.

H2O(l) → H2O(g)

Conduction (the transfer of energy through a material) is another example of a physical change. As energy is

transferred from one material to another, the energy of each material is changed, but not its chemical makeup.

Dissolving one substance in another is also a physical change.

Definition: Physical change

A change that can be seen or felt, but that doesn’t involve the break up of the particles in

the reaction. During a physical change, the form of matter may change, but not its identity.

A change in temperature is an example of a physical change.

You can think of a physical change as a person who is standing still. When they start to move (start walking) then

a change has occurred and this is similar to a physical change.

There are some important things to remember about physical changes in matter:

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1.56. CHEMICAL CHANGES IN MATTER CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE

1. Arrangement of particles

When a physical change occurs, the particles (e.g. atoms, molecules) may re-arrange themselves without

actually breaking up in any way. In the example of evaporation that we used earlier, the water molecules

move further apart as their temperature (and therefore energy) increases. The same would be true if ice

were to melt. In the solid phase, water molecules are packed close together in a very ordered way, but when

the ice is heated, the molecules overcome the forces holding them together and they move apart. Once

again, the particles have re-arranged themselves, but have not broken up.

H2O(s) → H2O(l) (8.1)

Figure 8.1 shows this more clearly. In each phase of water, the water molecule itself stays the same, but

the way the molecules are arranged has changed. Note that in the solid phase, we simply show the water

molecules as spheres. This makes it easier to see how tightly packed the molecules are. In reality the water

molecules would all look the same.

Figure 8.1: The arrangement of water molecules in the three phases of matter

2. Conservation of mass

In a physical change, the total mass, the number of atoms and the number of molecules will always stay

the same. In other words you will always have the same number of molecules or atoms at the end of the

change as you had at the beginning.

3. Energy changes

Energy changes may take place when there is a physical change in matter, but these energy changes are

normally smaller than the energy changes that take place during a chemical change.

4. Reversibility

Physical changes in matter are usually easier to reverse than chemical changes. Water vapour for example,

can be changed back to liquid water if the temperature is lowered. Liquid water can be changed into ice by

simply decreasing the temperature.

Activity: Physical change Use plastic pellets or marbles to represent water in the solid state. What do you need

to do to the pellets to represent the change from solid to liquid?

1.56 Chemical Changes in Matter

(section shortcode: C10054 )

When a chemical change takes place, new substances are formed in a chemical reaction. These new products

may have very different properties from the substances that were there at the start of the reaction.

The breakdown of copper (II) chloride to form copper and chlorine is an example of chemical change. A simplified

diagram of this reaction is shown in Figure 8.2. In this reaction, the initial substance is copper (II) chloride, but

once the reaction is complete, the products are copper and chlorine.

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CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE 1.56. CHEMICAL CHANGES IN MATTER

Figure 8.2: The decomposition of copper(II) chloride to form copper and chlorine. We write this as: CuCl2 →Cu + Cl2

Definition: Chemical change

The formation of new substances in a chemical reaction. One type of matter is changed

into something different.

There are some important things to remember about chemical changes:

1. Arrangement of particles

During a chemical change, the particles themselves are changed in some way. In the example of copper (II)

chloride that was used earlier, the CuCl2 molecules were split up into their component atoms. The number

of particles will change because each CuCl2 molecule breaks down into one copper atom (Cu) and one

chlorine molecule (CuCl2). However, what you should have noticed, is that the number of atoms of each

element stays the same, as does the total mass of the atoms. This will be discussed in more detail in a later

section.

2. Energy changes

The energy changes that take place during a chemical reaction are much greater than those that take place

during a physical change in matter. During a chemical reaction, energy is used up in order to break bonds,

and then energy is released when the new product is formed. This will be discussed in more detail in

"Energy changes in chemical reactions" (Section 7.4: Energy changes in chemical reactions).

3. Reversibility

Chemical changes are far more difficult to reverse than physical changes.

We will consider two types of chemical reactions: decomposition reactions and synthesis reactions.

1.56.1 Decomposition reactions

A decomposition reaction occurs when a chemical compound is broken down into elements or smaller com-

pounds. The generalised equation for a decomposition reaction is:

AB → A+ B

One example of such a reaction is the decomposition of mercury (II) oxide (Figure 8.3) to form mercury and

oxygen according to the following equation:

2HgO → 2Hg + O2 (8.2)

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1.56. CHEMICAL CHANGES IN MATTER CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE

Figure 8.3: The decomposition of HgO to form Hg and O2

The decomposition of hydrogen peroxide is another example.

Experiment : The decomposition of hydrogen peroxide

Aim:

To observe the decomposition of hydrogen peroxide when it is heated.

Apparatus:

Dilute hydrogen peroxide (about 3%); manganese dioxide; test tubes; a water bowl; stopper and delivery tube

Method:

1. Put a small amount (about 5 ml) of hydrogen peroxide in a test tube.

2. Set up the apparatus as shown in Figure 8.4

3. Very carefully add a small amount (about 0,5 g) of manganese dioxide to the test tube containing hydrogen

peroxide.

Results:

You should observe a gas bubbling up into the second test tube. This reaction happens quite rapidly.

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CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE 1.56. CHEMICAL CHANGES IN MATTER

Conclusions:

When hydrogen peroxide is added to manganese dioxide it decomposes to form oxygen and water. The chemical

decomposition reaction that takes place can be written as follows:

2H2O2 → 2H2O+O2 (8.3)

Note that the manganese dioxide is a catalyst and is not shown in the reaction. (A catalyst helps speed up a

chemical reaction.)

NOTE: The previous experiment used the downward displacement of water to collect

a gas. This is a very common way to collect a gas in chemistry. The oxygen that is

evolved in this reaction moves along the delivery tube and then collects in the top of

the test tube. It does this because it is lighter than water and so displaces the water

downwards. If you use a test tube with an outlet attached, you could collect the oxygen

into jars and store it for use in other experiments.

The above experiment can be very vigourous and produce a lot of oxygen very rapidly. For this reason you use

dilute hydrogen peroxide and only a small amount of manganese dioxide.

1.56.2 Synthesis reactions

During a synthesis reaction, a new product is formed from elements or smaller compounds. The generalised

equation for a synthesis reaction is as follows:

A+ B → AB (8.4)

One example of a synthesis reaction is the burning of magnesium in oxygen to form magnesium oxide(Figure 8.5).

The equation for the reaction is:

2Mg + O2 → 2MgO (8.5)

Figure 8.5: The synthesis of magnesium oxide (MgO) from magnesium and oxygen

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1.56. CHEMICAL CHANGES IN MATTER CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE

Experiment: Chemical reactions involving iron and sulphur

Aim:

To demonstrate the synthesis of iron sulphide from iron and sulphur.

Apparatus:

5,6 g iron filings and 3,2 g powdered sulphur; porcelain dish; test tube; Bunsen burner

Method:

1. Measure the quantity of iron and sulphur that you need and mix them in a porcelain dish.

2. Take some of this mixture and place it in the test tube. The test tube should be about 1/3 full.

3. This reaction should ideally take place in a fume cupboard. Heat the test tube containing the mixture over

the Bunsen burner. Increase the heat if no reaction takes place. Once the reaction begins, you will need to

remove the test tube from the flame. Record your observations.

4. Wait for the product to cool before breaking the test tube with a hammer. Make sure that the test tube is

rolled in paper before you do this, otherwise the glass will shatter everywhere and you may be hurt.

5. What does the product look like? Does it look anything like the original reactants? Does it have any of the

properties of the reactants (e.g. the magnetism of iron)?

Results:

1. After you removed the test tube from the flame, the mixture glowed a bright red colour. The reaction is

exothermic and produces energy.

2. The product, iron sulphide, is a dark colour and does not share any of the properties of the original reactants.

It is an entirely new product.

Conclusions:

A synthesis reaction has taken place. The equation for the reaction is:

Fe + S → FeS (8.6)

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CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE 1.57. ENERGY CHANGES IN CHEMICAL REACTIONS

Investigation : Physical or chemical change?

Apparatus: Bunsen burner, 4 test tubes, a test tube rack and a test tube holder, small spatula, pipette, magnet,

a birthday candle, NaCl (table salt), 0, 1M AgNO3, 6M HCl, magnesium ribbon, iron filings, sulphur.

WARNING: AgNO3 stains the skin. Be careful when working with it or use gloves.

Method:

1. Place a small amount of wax from a birthday candle into a test tube and heat it over the bunsen burner until

it melts. Leave it to cool.

2. Add a small spatula of NaCl to 5 ml water in a test tube and shake. Then use the pipette to add 10 drops of

AgNO3 to the sodium chloride solution. NOTE: Please be careful AgNO3 causes bad stains!!

3. Take a 5 cm piece of magnesium ribbon and tear it into 1 cm pieces. Place two of these pieces into a test

tube and add a few drops of 6M HCl. NOTE: Be very careful when you handle this acid because it can

cause major burns.

4. Take about 0,5 g iron filings and 0,5 g sulphur. Test each substance with a magnet. Mix the two samples in

a test tube and run a magnet alongside the outside of the test tube.

5. Now heat the test tube that contains the iron and sulphur. What changes do you see? What happens now,

if you run a magnet along the outside of the test tube?

6. In each of the above cases, record your observations.

Questions: Decide whether each of the following changes are physical or chemical and give a reason for your

answer in each case. Record your answers in the table below:

Description Physical or chemical change Reason

melting candle wax

dissolving NaCl

mixing NaCl with AgNO3

tearing magnesium ribbon

adding HCl to magnesium ribbon

mixing iron and sulphur

heating iron and sulphur

Table 8.1

1.57 Energy changes in chemical reactions

(section shortcode: C10055 )

All reactions involve some change in energy. During a physical change in matter, such as the evaporation of

liquid water to water vapour, the energy of the water molecules increases. However, the change in energy is

much smaller than in chemical reactions.

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1.58. CONSERVATION OF ATOMS AND MASS IN REACTIONSCHAPTER 1. PHYSICAL AND CHEMICAL CHANGE

When a chemical reaction occurs, some bonds will break, while new bonds may form. Energy changes in

chemical reactions result from the breaking and forming of bonds. For bonds to break, energy must be absorbed.

When new bonds form, energy will be released because the new product has a lower energy than the ‘in between’

stage of the reaction when the bonds in the reactants have just been broken.

In some reactions, the energy that must be absorbed to break the bonds in the reactants is less than the total

energy that is released when new bonds are formed. This means that in the overall reaction, energy is released.

This type of reaction is known as an exothermic reaction. In other reactions, the energy that must be absorbed

to break the bonds in the reactants is more than the total energy that is released when new bonds are formed.

This means that in the overall reaction, energy must be absorbed from the surroundings. This type of reaction is

known as an endothermic reaction. Most decomposition reactions are endothermic and heating is needed for

the reaction to occur. Most synthesis reactions are exothermic, meaning that energy is given off in the form of

heat or light.

More simply, we can describe the energy changes that take place during a chemical reaction as:

Total energy absorbed to break bonds - Total energy released when new bonds form

So, for example, in the reaction...

2Mg + O2 → 2MgO

Energy is needed to break the O−O bonds in the oxygen molecule so that new Mg −O bonds can be formed,

and energy is released when the product (MgO) forms.

Despite all the energy changes that seem to take place during reactions, it is important to remember that energy

cannot be created or destroyed. Energy that enters a system will have come from the surrounding environment

and energy that leaves a system will again become part of that environment. This is known as the conservation

of energy principle.

Definition: Conservation of energy principle

Energy cannot be created or destroyed. It can only be changed from one form to another.

Chemical reactions may produce some very visible and often violent changes. An explosion, for example, is a

sudden increase in volume and release of energy when high temperatures are generated and gases are released.

For example, NH4NO3 can be heated to generate nitrous oxide. Under these conditions, it is highly sensitive and

can detonate easily in an explosive exothermic reaction.

1.58 Conservation of atoms and mass in reactions

(section shortcode: C10056 )

The total mass of all the substances taking part in a chemical reaction is conserved during a chemical reaction.

This is known as the law of conservation of mass. The total number of atoms of each element also remains

the same during a reaction, although these may be arranged differently in the products.

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CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE1.58. CONSERVATION OF ATOMS AND MASS IN REACTIONS

We will use two of our earlier examples of chemical reactions to demonstrate this:

1. The decomposition of hydrogen peroxide into water and oxygen

2H2O2 → 2H2O+O2

Left hand side of the equation

Total atomic mass = (4× 1) + (4× 16) = 68 u

Number of atoms of each element = (4×H) + (4×O)

Right hand side of the equation

Total atomic mass = (4× 1) + (4× 16) = 68 u

Number of atoms of each element = (4×H) + (4×O)

Both the atomic mass and the number of atoms of each element are conserved in the reaction.

2. The synthesis of magnesium and oxygen to form magnesium oxide

2Mg + O2 → 2MgO (8.7)

Left hand side of the equation

Total atomic mass = (2× 24, 3) + (2× 16) = 80, 6 u

Number of atoms of each element = (2×Mg) + (2×O)

Right hand side of the equation

Total atomic mass = (2× 24, 3) + (2× 16) = 80, 6 u

Number of atoms of each element = (2×Mg) + (2×O)

Both the atomic mass and the number of atoms of each element are conserved in the reaction.

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1.58. CONSERVATION OF ATOMS AND MASS IN REACTIONSCHAPTER 1. PHYSICAL AND CHEMICAL CHANGE

1.58.1 Activity : The conservation of atoms in chemical reactions

Materials:

1. Coloured marbles or small balls to represent atoms. Each colour will represent a different element.

2. Prestik

Method:

1. Choose a reaction from any that have been used in this chapter or any other balanced chemical reaction

that you can think of. To help to explain this activity, we will use the decomposition reaction of calcium

carbonate to produce carbon dioxide and calcium oxide. CaCO3 → CO2 +CaO2. Stick marbles together to represent the reactants and put these on one side of your table. In this example

you may for example join one red marble (calcium), one green marble (carbon) and three yellow marbles

(oxygen) together to form the molecule calcium carbonate (CaCO3).

3. Leaving your reactants on the table, use marbles to make the product molecules and place these on the

other side of the table.

4. Now count the number of atoms on each side of the table. What do you notice?

5. Observe whether there is any difference between the molecules in the reactants and the molecules in the

products.

Discussion You should have noticed that the number of atoms in the reactants is the same as the number of

atoms in the product. The number of atoms is conserved during the reaction. However, you will also see that the

molecules in the reactants and products is not the same. The arrangement of atoms is not conserved during the

reaction.

1.58.2 Experiment: Conservation of matter

Aim: To prove the law of conservation of matter experimentally.

Materials: Test tubes; glass beaker; lead (II) nitrate; sodium iodide; hydrochloric acid; bromothymol blue; Cal-C-

Vita tablet, plastic bag; rubber band; mass meter

Method: Reaction 1

1. Carefully weigh out 5 g of lead (II) nitrate.

2. Dissolve the lead nitrate in 100 ml of water.

3. Weigh the lead nitrate solution.

4. Weigh out 4,5 g of sodium iodide and dissolve this in the lead (II) nitrate solution.

5. Weigh the beaker containing the lead nitrate and sodium iodide mixture.

Reaction 2

1. Measure out 20 ml of sodium hydroxide.

2. Add a few drops of bromothymol blue to the sodium hydroxide.

3. Weigh the sodium hydroxide.

4. Weigh 5 ml of hydrochloric acid.

5. Add 5 ml of hydrochloric acid to the sodium hydroxide. Repeat this step until you observe a colour change

(this should occur around 20 ml).

6. Weigh the final solution.

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CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE 1.59. LAW OF CONSTANT COMPOSITION

Reaction 3

1. Measure out 100 ml of water into a beaker.

2. Weigh the beaker with water in it.

3. Place the Cal-C-Vita tablet into the plastic bag.

4. Weigh the Cal-C-Vita tablet and the plastic bag.

5. Place the plastic bag over the beaker, being careful to not let the tablet fall into the water

6. Seal the bag around the beaker using the rubber band. Drop the tablet into the water.

7. Observe what happens.

8. Weigh the bag and beaker containing the solution.

Results: Fill in the following table for reactants (starting materials) and products (ending materials) masses.

For the second reaction, you will simply take the mass of 5 ml of hydrochloric acid and multiply it by how many

amounts you put in, for example, if you put 4 amounts in, then you would have 20 ml and 4 times the mass of 5

ml.

Reaction 1 Reaction 2 Reaction 3

Reactants

Products

Table 8.2

Add the masses for the reactants for each reaction. Do the same for the products. For each reaction compare

the mass of the reactants to the mass of the products. What do you notice? Is the mass conserved?

In the experiment above you should have found that the mass at the start of the reaction is the same as the mass

at the end of the reaction. You may have found that these masses differed slightly, but this is due to errors in

measurements and in performing experiments (all scientists make some errors in performing experiments).

1.59 Law of constant composition

(section shortcode: C10057 )

In any given chemical compound, the elements always combine in the same proportion with each other. This is

the law of constant proportion.

The law of constant composition says that, in any particular chemical compound, all samples of that compound

will be made up of the same elements in the same proportion or ratio. For example, any water molecule is always

made up of two hydrogen atoms and one oxygen atom in a 2:1 ratio. If we look at the relative masses of oxygen

and hydrogen in a water molecule, we see that 94% of the mass of a water molecule is accounted for by oxygen

and the remaining 6% is the mass of hydrogen. This mass proportion will be the same for any water molecule.

This does not mean that hydrogen and oxygen always combine in a 2:1 ratio to form H2O. Multiple proportions

are possible. For example, hydrogen and oxygen may combine in different proportions to form H2O2 rather than

H2O. In H2O2, the H:O ratio is 1:1 and the mass ratio of hydrogen to oxygen is 1:16. This will be the same for

any molecule of hydrogen peroxide.

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1.60. VOLUME RELATIONSHIPS IN GASES CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE

1.60 Volume relationships in gases

(section shortcode: C10058 )

In a chemical reaction between gases, the relative volumes of the gases in the reaction are present in a ratio of

small whole numbers if all the gases are at the same temperature and pressure. This relationship is also known

as Gay-Lussac’s Law.

For example, in the reaction between hydrogen and oxygen to produce water, two volumes of H2 react with 1

volume of O2 to produce 2 volumes of H2O.

2H2 +O2 → 2H2O

In the reaction to produce ammonia, one volume of nitrogen gas reacts with three volumes of hydrogen gas to

produce two volumes of ammonia gas.

N2 + 3H2 → 2NH3

This relationship will also be true for all other chemical reactions.

1.61 Summary

(section shortcode: C10059 )

The following video provides a summary of the concepts covered in this chapter.

Physical and chemical change (Video: P10060 )

1. Matter does not stay the same. It may undergo physical or chemical changes.

2. A physical change means that the form of matter may change, but not its identity. For example, when

water evaporates, the energy and the arrangement of water molecules will change, but not the structure of

the water molecules themselves.

3. During a physical change, the arrangement of particles may change but the mass, number of atoms and

number of molecules will stay the same.

4. Physical changes involve small changes in energy and are easily reversible.

5. A chemical change occurs when one or more substances change into other materials. A chemical reaction

involves the formation of new substances with different properties. For example, magnesium and oxygen

react to form magnesium oxide (MgO)

6. A chemical change may involve a decomposition or synthesis reaction. During chemical change, the

mass and number of atoms is conserved, but the number of molecules is not always the same.

7. Chemical reactions involve larger changes in energy. During a reaction, energy is needed to break bonds

in the reactants and energy is released when new products form. If the energy released is greater than the

energy absorbed, then the reaction is exothermic. If the energy released is less than the energy absorbed,

then the reaction is endothermic. Chemical reactions are not easily reversible.

8. Decomposition reactions are usually endothermic and synthesis reactions are usually exothermic.

9. The law of conservation of mass states that the total mass of all the substances taking part in a chemical

reaction is conserved and the number of atoms of each element in the reaction does not change when a

new product is formed.

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CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE 1.61. SUMMARY

10. The conservation of energy principle states that energy cannot be created or destroyed, it can only

change from one form to another.

11. The law of constant composition states that in any particular compound, all samples of that compound

will be made up of the same elements in the same proportion or ratio.

12. Gay-Lussac’s Law states that in a chemical reaction between gases, the relative volumes of the gases in

the reaction are present in a ratio of small whole numbers if all the gases are at the same temperature and

pressure.

1.61.1 End of chapter exercises

1. For each of the following definitions give one word or term:

a. A change that can be seen or felt, where the particles involved are not broken up in any way

b. The formation of new substances in a chemical reaction

c. A reaction where a new product is formed from elements or smaller compounds

2. State the conservation of energy principle.

3. Explain how a chemical change differs from a physical change.

4. Complete the following table by saying whether each of the descriptions is an example of a physical or

chemical change:

Description Physical or chemical

hot and cold water mix together

milk turns sour

a car starts to rust

food digests in the stomach

alcohol disappears when it is placed on your skin

warming food in a microwave

separating sand and gravel

fireworks exploding

Table 8.3

5. For each of the following reactions, say whether it is an example of a synthesis or decomposition reaction:

a. (NH4)2 CO3 → NH3 +CO2 +H2Ob. N2 (g) + 3H2 (g) → 2NH3

c. CaCO3 (s) → CaO + CO2

6. For the following equation: CaCO3 (s) → CaO + CO2 show that the ’law of conservation of mass’ applies.

Find the answers with the shortcodes:

(1.) l2z (2.) l2u (3.) l2J (4.) l3q (5.) l3l (6.) l3i

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1.61. SUMMARY CHAPTER 1. PHYSICAL AND CHEMICAL CHANGE

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Representing Chemical Change

1.62 Introduction

(section shortcode: C10061 )

As we have already mentioned, a number of changes can occur when elements react with one another. These

changes may either be physical or chemical. One way of representing these changes is through balanced

chemical equations. A chemical equation describes a chemical reaction by using symbols for the elements

involved. For example, if we look at the reaction between iron (Fe) and sulphur (S) to form iron sulphide (FeS), we

could represent these changes either in words or using chemical symbols:

iron + sulphur → iron sulphide

or

Fe + S → FeS

Another example would be:

ammonia + oxygen → nitric oxide + water

or

4NH3 + 5O2 → 4NO + 6H2O

Compounds on the left of the arrow are called the reactants and these are needed for the reaction to take place.

In this equation, the reactants are ammonia and oxygen. The compounds on the right are called the products

and these are what is formed from the reaction.

In order to be able to write a balanced chemical equation, there are a number of important things that need to be

done:

1. Know the chemical symbols for the elements involved in the reaction

2. Be able to write the chemical formulae for different reactants and products

3. Balance chemical equations by understanding the laws that govern chemical change

4. Know the state symbols for the equation

We will look at each of these steps separately in the next sections.

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1.63. CHEMICAL SYMBOLS CHAPTER 1. REPRESENTING CHEMICAL CHANGE

1.63 Chemical symbols

(section shortcode: C10062 )

It is very important to know the chemical symbols for common elements in the Periodic Table, so that you are

able to write chemical equations and to recognise different compounds.

1.63.1 Revising common chemical symbols

• Write down the chemical symbols and names of all the elements that you know.

• Compare your list with another learner and add any symbols and names that you don’t have.

• Spend some time, either in class or at home, learning the symbols for at least the first twenty elements in

the periodic table. You should also learn the symbols for other common elements that are not in the first

twenty.

• Write a short test for someone else in the class and then exchange tests with them so that you each have

the chance to answer one.

1.64 Writing chemical formulae

(section shortcode: C10063 )

A chemical formula is a concise way of giving information about the atoms that make up a particular chemical

compound. A chemical formula shows each element by its symbol and also shows how many atoms of each

element are found in that compound. The number of atoms (if greater than one) is shown as a subscript.

Examples: CH4 (methane)

Number of atoms: (1× carbon) + (4× hydrogen) = 5 atoms in one methane molecule

H2SO4 (sulphuric acid)

Number of atoms: (2× hydrogen) + (1× sulphur) + (4× oxygen) = 7 atoms in one molecule of sulphuric acid

A chemical formula may also give information about how the atoms are arranged in a molecule if it is written

in a particular way. A molecule of ethane, for example, has the chemical formula C2H6. This formula tells us

how many atoms of each element are in the molecule, but doesn’t tell us anything about how these atoms are

arranged. In fact, each carbon atom in the ethane molecule is bonded to three hydrogen atoms. Another way

of writing the formula for ethane is CH3CH3. The number of atoms of each element has not changed, but this

formula gives us more information about how the atoms are arranged in relation to each other.

The slightly tricky part of writing chemical formulae comes when you have to work out the ratio in which the

elements combine. For example, you may know that sodium (Na) and chlorine (Cl) react to form sodium chloride,

but how do you know that in each molecule of sodium chloride there is only one atom of sodium for every one

atom of chlorine? It all comes down to the valency of an atom or group of atoms. Valency is the number of bonds

that an element can form with another element. Working out the chemical formulae of chemical compounds using

their valency, will be covered in Grade 11. For now, we will use formulae that you already know.

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CHAPTER 1. REPRESENTING CHEMICAL CHANGE 1.65. BALANCING CHEMICAL EQUATIONS

1.65 Balancing chemical equations

(section shortcode: C10064 )

1.65.1 The law of conservation of mass

In order to balance a chemical equation, it is important to understand the law of conservation of mass.

Definition: The law of conservation of mass

The mass of a closed system of substances will remain constant, regardless of the pro-

cesses acting inside the system. Matter can change form, but cannot be created or de-

stroyed. For any chemical process in a closed system, the mass of the reactants must

equal the mass of the products.

In a chemical equation then, the mass of the reactants must be equal to the mass of the products. In order to

make sure that this is the case, the number of atoms of each element in the reactants must be equal to the

number of atoms of those same elements in the products. Some examples are shown below:

Example 1:

Fe + S → FeS (9.1)

Reactants

Atomic mass of reactants = 55, 8 u+ 32, 1 u = 87, 9 u

Number of atoms of each element in the reactants: (1× Fe) and (1× S)

Products

Atomic mass of products = 55, 8 u+ 32, 1 u = 87, 9 u

Number of atoms of each element in the products: (1× Fe) and (1× S)

Since the number of atoms of each element is the same in the reactants and in the products, we say that the

equation is balanced.

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1.65. BALANCING CHEMICAL EQUATIONS CHAPTER 1. REPRESENTING CHEMICAL CHANGE

Example 2:

H2 +O2 → H2O (9.2)

Reactants

Atomic mass of reactants = (1 + 1) + (16 + 16) = 34 u

Number of atoms of each element in the reactants: (2×H) and (2×O)

Product

Atomic mass of product = (1 + 1 + 16) = 18 u

Number of atoms of each element in the product: (2×H) and (1×O)

Since the total atomic mass of the reactants and the products is not the same and since there are more oxygen

atoms in the reactants than there are in the product, the equation is not balanced.

Example 3:

NaOH+HCl → NaCl + H2O (9.3)

Reactants

Atomic mass of reactants = (23 + 6 + 1) + (1 + 35, 4) = 76, 4 u

Number of atoms of each element in the reactants: (1×Na) + (1×O) + (2×H) + (1× Cl)

Products

Atomic mass of products = (23 + 35, 4) + (1 + 1 + 16) = 76, 4 u

Number of atoms of each element in the products: (1×Na) + (1×O) + (2×H) + (1× Cl)

Since the number of atoms of each element is the same in the reactants and in the products, we say that the

equation is balanced.

We now need to find a way to balance those equations that are not balanced so that the number of atoms of each

element in the reactants is the same as that for the products. This can be done by changing the coefficients of

the molecules until the atoms on each side of the arrow are balanced. You will see later that these coefficients

tell us something about the mole ratio in which substances react. They also tell us about the volume relationship

between gases in the reactants and products.

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CHAPTER 1. REPRESENTING CHEMICAL CHANGE 1.65. BALANCING CHEMICAL EQUATIONS

TIP: Coefficients : Remember that if you put a number in front of a molecule, that

number applies to the whole molecule. For example, if you write 2H2O, this means

that there are 2 molecules of water. In other words, there are 4 hydrogen atoms and 2

oxygen atoms. If we write 3HCl, this means that there are 3 molecules of HCl. In other

words there are 3 hydrogen atoms and 3 chlorine atoms in total. In the first example,

2 is the coefficient and in the second example, 3 is the coefficient.

1.65.2 Activity: Balancing chemical equations

You will need: coloured balls (or marbles), prestik, a sheet of paper and coloured pens.

We will try to balance the following equation:

Al + O2 → Al2O3 (9.4)

Take 1 ball of one colour. This represents a molecule of Al. Take two balls of another colour and stick them

together. This represents a molecule of O2. Place these molecules on your left. Now take two balls of one colour

and three balls of another colour to form a molecule of Al2O3. Place these molecules on your right. On a piece

of paper draw coloured circles to represent the balls. Draw a line down the center of the paper to represent the

molecules on the left and on the right.

Count the number of balls on the left and the number on the right. Do you have the same number of each colour

on both sides? If not the equation is not balanced. How many balls will you have to add to each side to make the

number of balls the same? How would you add these balls?

You should find that you need 4 balls of one colour for Al and 3 pairs of balls of another colour (i.e. 6 balls in

total) for O2 on the left side. On the right side you should find that you need 2 clusters of balls for Al2O3. We say

that the balanced equation is:

4Al + 3O2 → 2Al2O3 (9.5)

Repeat this process for the following reactions:

• CH4 + 2O2 → CO2 + 2H2O• 2H2 +O2 → 2H2O• Zn + 2HCl → ZnCl2 +H2

1.65.3 Steps to balance a chemical equation

When balancing a chemical equation, there are a number of steps that need to be followed.

• STEP 1: Identify the reactants and the products in the reaction and write their chemical formulae.

• STEP 2: Write the equation by putting the reactants on the left of the arrow and the products on the right.

• STEP 3: Count the number of atoms of each element in the reactants and the number of atoms of each

element in the products.

• STEP 4: If the equation is not balanced, change the coefficients of the molecules until the number of atoms

of each element on either side of the equation balance.

• STEP 5: Check that the atoms are in fact balanced.

• STEP 6 (we will look at this a little later): Add any extra details to the equation e.g. phase.

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1.65. BALANCING CHEMICAL EQUATIONS CHAPTER 1. REPRESENTING CHEMICAL CHANGE

Exercise 9.1: Balancing chemical equations 1 Balance the following

equation:

Mg +HCl → MgCl2 +H2 (9.6)

Solution to Exercise

Step 1. Reactants: Mg = 1 atom; H = 1 atom and Cl = 1 atomProducts: Mg = 1 atom; H = 2 atoms and Cl = 2 atoms

Step 2. The equation is not balanced since there are 2 chlorine atoms in the

product and only 1 in the reactants. If we add a coefficient of 2 to

the HCl to increase the number of H and Cl atoms in the reactants,

the equation will look like this:

Mg + 2HCl → MgCl2 +H2 (9.7)

Step 3. If we count the atoms on each side of the equation, we find the

following:

Reactants: Mg = 1 atom; H = 2 atom and Cl = 2 atomProducts: Mg = 1 atom; H = 2 atom and Cl = 2 atomThe equation is balanced. The final equation is:

Mg + 2HCl → MgCl2 +H2 (9.8)

Exercise 9.2: Balancing chemical equations 2 Balance the following

equation:

CH4 +O2 → CO2 +H2O (9.9)

Solution to Exercise

Step 1. Reactants: C = 1; H = 4 and O = 2Products: C = 1; H = 2 and O = 3

Step 2. If we add a coefficient of 2 to H2O, then the number of hydrogen

atoms in the reactants will be 4, which is the same as for the reac-

tants. The equation will be:

CH4 +O2 → CO2 + 2H2O (9.10)

Step 3. Reactants: C = 1; H = 4 and O = 2Products: C = 1; H = 4 and O = 4

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You will see that, although the number of hydrogen atoms now bal-

ances, there are more oxygen atoms in the products. You now need

to repeat the previous step. If we put a coefficient of 2 in front of O2,

then we will increase the number of oxygen atoms in the reactants

by 2. The new equation is:

CH4 + 2O2 → CO2 + 2H2O (9.11)

When we check the number of atoms again, we find that the num-

ber of atoms of each element in the reactants is the same as the

number in the products. The equation is now balanced.

Exercise 9.3: Balancing chemical equations 3 Nitrogen gas reacts

with hydrogen gas to form ammonia. Write a balanced chemical equation

for this reaction.

Solution to Exercise

Step 1. The reactants are nitrogen (N2) and hydrogen (H2) and the product

is ammonia (NH3).

Step 2. The equation is as follows:

N2 +H2 → NH3 (9.12)

Step 3. Reactants: N = 2 and H = 2Products: N = 1 and H = 3

Step 4. In order to balance the number of nitrogen atoms, we could rewrite

the equation as:

N2 +H2 → 2NH3 (9.13)

Step 5. In the above equation, the nitrogen atoms now balance, but the

hydrogen atoms don’t (there are 2 hydrogen atoms in the reactants

and 6 in the product). If we put a coefficient of 3 in front of the

hydrogen (H2), then the hydrogen atoms and the nitrogen atoms

balance. The final equation is:

N2 + 3H2 → 2NH3 (9.14)

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Exercise 9.4: Balancing chemical equations 4 In our bodies, sugar

(C6H12O6) reacts with the oxygen we breathe in to produce carbon diox-

ide, water and energy. Write the balanced equation for this reaction.

Solution to Exercise

Step 1. Reactants: sugar (C6H12O6) and oxygen (O2)

Products: carbon dioxide (CO2) and water (H2O)

Step 2.

C6H12O6 +O2 → CO2 +H2O (9.15)

Step 3. Reactants: C = 6; H = 12 and O = 8Products: C = 1; H = 2 and O = 3

Step 4. It is easier to start with carbon as it only appears once on each side.

If we add a 6 in front of CO2, the equation looks like this:

C6H12O6 +O2 → 6CO2 +H2O (9.16)

Reactants: C = 6; H = 12 and O = 8Products: C = 6; H = 2 and O = 13

Step 5. Let’s try to get the number of hydrogens the same this time.

C6H12O6 +O2 → 6CO2 + 6H2O (9.17)

Reactants: C = 6; H = 12 and O = 8Products: C = 6; H = 12 and O = 18

Step 6.

C6H12O6 + 12O2 → 6CO2 + 6H2O (9.18)

Reactants: C = 6; H = 12 and O = 18Products: C = 6; H = 12 and O = 18

This simulation allows you to practice balancing simple equations.

(Simulation: lbD)

Balancing simple chemical equations

Balance the following equations:

1. Hydrogen fuel cells are extremely important in the development of alternative energy sources. Many of these

cells work by reacting hydrogen and oxygen gases together to form water, a reaction which also produces

electricity. Balance the following equation: H2 (g) + O2 (g) → H2O(l)

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2. The synthesis of ammonia (NH3), made famous by the German chemist Fritz Haber in the early 20th

century, is one of the most important reactions in the chemical industry. Balance the following equation

used to produce ammonia: N2 (g) + H2 (g) → NH3 (g)3. Mg + P4 → Mg3P2

4. Ca + H2O → Ca (OH)2 +H2

5. CuCO3 +H2SO4 → CuSO4 +H2O+ CO2

6. CaCl2 +Na2CO3 → CaCO3 +NaCl7. C12H22O11 +O2 → H2O+ CO2

8. Barium chloride reacts with sulphuric acid to produce barium sulphate and hydrochloric acid.

9. Ethane (C2H6) reacts with oxygen to form carbon dioxide and steam.

10. Ammonium carbonate is often used as a smelling salt. Balance the following reaction for the decomposition

of ammonium carbonate: (NH4) 2CO3 (s) → NH3 (aq) + CO2 (g) + H2O(l)

Find the answers with the shortcodes:

(1.) lOg (2.) lO4 (3.) lO2 (4.) lOT (5.) lOb (6.) lOj

(7.) lOD (8.) lOW (9.) lOZ (10.) lOB

1.66 State symbols and other information

(section shortcode: C10065 )

The state (phase) of the compounds can be expressed in the chemical equation. This is done by placing the

correct label on the right hand side of the formula. There are only four labels that can be used:

1. (g) for gaseous compounds

2. (l) for liquids

3. (s) for solid compounds

4. (aq) for an aqueous (water) solution

Occasionally, a catalyst is added to the reaction. A catalyst is a substance that speeds up the reaction without

undergoing any change to itself. In a chemical equation, this is shown by using the symbol of the catalyst above

the arrow in the equation.

To show that heat is needed for a reaction, a Greek delta (∆) is placed above the arrow in the same way as the

catalyst.

TIP: You may remember from Physical and chemical change (Chapter 7) that energy

cannot be created or destroyed during a chemical reaction but it may change form. In

an exothermic reaction, ∆H is less than zero and in an endothermic reaction, ∆H is

greater than zero. This value is often written at the end of a chemical equation.

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1.66. STATE SYMBOLS AND OTHER INFORMATION CHAPTER 1. REPRESENTING CHEMICAL CHANGE

Exercise 9.5: Balancing chemical equations 5 Solid zinc metal re-

acts with aqueous hydrochloric acid to form an aqueous solution of zinc

chloride (ZnCl2)and hydrogen gas. Write a balanced equation for this

reaction.

Solution to Exercise

Step 1. The reactants are zinc (Zn) and hydrochloric acid (HCl). The prod-

ucts are zinc chloride (ZnCl2) and hydrogen (H2).

Step 2.

Zn + HCl → ZnCl2 +H2 (9.19)

Step 3. You will notice that the zinc atoms balance but the chlorine and

hydrogen atoms don’t. Since there are two chlorine atoms on the

right and only one on the left, we will give HCl a coefficient of 2 so

that there will be two chlorine atoms on each side of the equation.

Zn + 2HCl → ZnCl2 +H2 (9.20)

Step 4. When you look at the equation again, you will see that all the atoms

are now balanced.

Step 5. In the initial description, you were told that zinc was a metal, hy-

drochloric acid and zinc chloride were in aqueous solutions and

hydrogen was a gas.

Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g) (9.21)

Exercise 9.6: Balancing chemical equations 5 (advanced) Balance

the following equation:

(NH4) 2SO4 +NaOH → NH3 +H2O+Na2SO4 (9.22)

In this example, the first two steps are not necessary because the reac-

tants and products have already been given.

Solution to Exercise

Step 1. With a complex equation, it is always best to start with atoms that

appear only once on each side i.e. Na, N and S atoms. Since the

S atoms already balance, we will start with Na and N atoms. There

are two Na atoms on the right and one on the left. We will add a

second Na atom by giving NaOH a coefficient of two. There are two

N atoms on the left and one on the right. To balance the N atoms,

NH3 will be given a coefficient of two. The equation now looks as

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CHAPTER 1. REPRESENTING CHEMICAL CHANGE 1.66. STATE SYMBOLS AND OTHER INFORMATION

follows:

(NH4) 2SO4 + 2NaOH → 2NH3 +H2O+Na2SO4 (9.23)

Step 2. N, Na and S atoms balance, but O and H atoms do not. There are

six O atoms and ten H atoms on the left, and five O atoms and

eight H atoms on the right. We need to add one O atom and two Hatoms on the right to balance the equation. This is done by adding

another H2O molecule on the right hand side. We now need to

check the equation again:

(NH4) 2SO4 + 2NaOH → 2NH3 + 2H2O+Na2SO4 (9.24)

The equation is now balanced.

The following video explains some of the concepts of balancing chemical equations.

Khan Academy video on balancing chemical equations (Video: P10066 )

1.66.1 Balancing more advanced chemical equations

Write balanced equations for each of the following reactions:

1. Al2O3 (s) + H2SO4 (aq) → Al2 (SO4) 3 (aq) + 3H2O (l)2. Mg (OH)2 (aq) + HNO3 (aq) → Mg (NO3) 2 (aq) + 2H2O (l)3. Lead (II) nitrate solution reacts with potassium iodide solution.

4. When heated, aluminium reacts with solid copper oxide to produce copper metal and aluminium oxide

(Al2O3).

5. When calcium chloride solution is mixed with silver nitrate solution, a white precipitate (solid) of silver chlo-

ride appears. Calcium nitrate (Ca (NO3) 2) is also produced in the solution.

Find the answers with the shortcodes:

(1.) lOK

Balanced equations are very important in chemistry. It is only by working with the balanced equations that

chemists can perform many different calculations that tell them what quantity of something reacts. In a later

chapter we will learn how to work with some of these calculations. We can interpret balanced chemical equations

in terms of the conservation of matter, the conservation of mass or the conservation of energy.

(Presentation: P10067 )

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1.67. SUMMARY CHAPTER 1. REPRESENTING CHEMICAL CHANGE

1.67 Summary

(section shortcode: C10068 )

• A chemical equation uses symbols to describe a chemical reaction.

• In a chemical equation, reactants are written on the left hand side of the equation and the products on the

right. The arrow is used to show the direction of the reaction.

• When representing chemical change, it is important to be able to write the chemical formula of a compound.

• In any chemical reaction, the law of conservation of mass applies. This means that the total atomic

mass of the reactants must be the same as the total atomic mass of the products. This also means that

the number of atoms of each element in the reactants must be the same as the number of atoms of each

element in the product.

• If the number of atoms of each element in the reactants is the same as the number of atoms of each element

in the product, then the equation is balanced.

• If the number of atoms of each element in the reactants is not the same as the number of atoms of each

element in the product, then the equation is not balanced.

• In order to balance an equation, coefficients can be placed in front of the reactants and products until the

number of atoms of each element is the same on both sides of the equation.

1.67.1 End of chapter exercises

1. Propane is a fuel that is commonly used as a heat source for engines and homes. Balance the following

equation for the combustion of propane: C3H8 (l) + O2 (g) → CO2 (g) + H2O(l)2. Aspartame, an artificial sweetener, has the formula C14H18N2O2. Write the balanced equation for its com-

bustion (reaction with O2) to form CO2 gas, liquid H2O, and N2 gas.

3. Fe2 (SO4) 3 +K(SCN) → K3Fe (SCN)6 +K2SO4

4. Chemical weapons were banned by the Geneva Protocol in 1925. According to this protocol, all chemicals

that release suffocating and poisonous gases are not to be used as weapons. White phosphorus, a very

reactive allotrope of phosphorus, was recently used during a military attack. Phosphorus burns vigorously

in oxygen. Many people got severe burns and some died as a result. The equation for this spontaneous

reaction is: P4 (s) + O2 (g) → P2O5 (s)

a. Balance the chemical equation.

b. Prove that the law of conservation of mass is obeyed during this chemical reaction.

c. Name the product formed during this reaction.

d. Classify the reaction as endothermic or exothermic. Give a reason for your answer.

e. Classify the reaction as a synthesis or decomposition reaction. Give a reason for your answer.

(DoE Exemplar Paper 2 2007)

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5. Mixing bleach (NaOCl) and ammonia (two common household cleaners) is very dangerous. When these

two substances are mixed they produce toxic chloaramine (NH2Cl) fumes. Balance the following equations

that occur when bleach and ammonia are mixed:

a. NaOCl (aq) + NH3 (aq) → NaONH3 (aq) + Cl2 (g)b. If there is more bleach than ammonia the following may occur: NaOCl + NH3 → NaOH+NCl3

Nitrogen trichloride (NCl3) is highly explosive.

c. If there is more ammonia than bleach the following may occur: NH3 +NaOCl → NaOH+NH2ClThese two products then react with ammonia as follows:

NH3 +NH2Cl + NaOH → N2H4 +NaCl + H2OOne last reaction occurs to stabilise the hydrazine and chloramine: NH2Cl + N2H4 → NH4Cl + N2

This reaction is highly exothermic and will explode.

6. Balance the following chemical equation: N2O5 → NO2 +O2

7. Sulphur can be produced by the Claus process. This two-step process involves reacting hydrogen sulphide

with oxygen and then reacting the sulphur dioxide that is produced with more hydrogen sulphide. The

equations for these two reactions are:

H2S + O2 → SO2 +H2O

H2S + SO2 → S + H2O(9.25)

Balance these two equations.

Find the answers with the shortcodes:

(1.) lOk (2.) lO0 (3.) lO8 (4.) lO9 (5.) l2S (6.) l2h

(7.) lTq

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Reactions in Aqueous Solutions

1.68 Introduction

(section shortcode: C10069 )

Many reactions in chemistry and all biological reactions (reactions in living systems) take place in water. We say

that these reactions take place in aqueous solution. Water has many unique properties and is plentiful on Earth.

For these reasons reactions in aqueous solutions occur frequently. In this chapter we will look at some of these

reactions in detail. Almost all the reactions that occur in aqueous solutions involve ions. We will look at three main

types of reactions that occur in aqueous solutions, namely precipitation reactions, acid-base reactions and redox

reactions. Before we can learn about the types of reactions, we need to first look at ions in aqueous solutions

and electrical conductivity.

1.69 Ions in aqueous solution

(section shortcode: C10070 )

Water is seldom pure. Because of the structure of the water molecule, substances can dissolve easily in it. This

is very important because if water wasn’t able to do this, life would not be able to survive. In rivers and the oceans

for example, dissolved oxygen means that organisms (such as fish) are still able to respire (breathe). For plants,

dissolved nutrients are also available. In the human body, water is able to carry dissolved substances from one

part of the body to another.

Many of the substances that dissolve are ionic and when they dissolve they form ions in solution. We are going

to look at how water is able to dissolve ionic compounds, how these ions maintain a balance in the human body,

how they affect water hardness and how they cause acid rain.

1.69.1 Dissociation in water

Water is a polar molecule (Figure 10.1). This means that one part of the molecule has a slightly positive charge

(positive pole) and the other part has a slightly negative charge (negative pole).

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Figure 10.1: Water is a polar molecule

It is the polar nature of water that allows ionic compounds to dissolve in it. In the case of sodium chloride (NaCl)for example, the positive sodium ions (Na+) will be attracted to the negative pole of the water molecule, while

the negative chloride ions (Cl−) will be attracted to the positive pole of the water molecule. In the process, the

ionic bonds between the sodium and chloride ions are weakened and the water molecules are able to work their

way between the individual ions, surrounding them and slowly dissolving the compound. This process is called

dissociation. A simplified representation of this is shown in Figure 10.2. We say that dissolution of a substance

has occurred when a substance dissociates or dissolves.

Definition: Dissociation

Dissociation in chemistry and biochemistry is a general process in which ionic compounds

separate or split into smaller molecules or ions, usually in a reversible manner.

Figure 10.2: Sodium chloride dissolves in water

The dissolution of sodium chloride can be represented by the following equation:

NaCl (s) → Na+(aq) + Cl−(aq)

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The symbols s (solid), l (liquid), g (gas) and aq (material is dissolved in water) are written after the chemical

formula to show the state or phase of the material. The dissolution of potassium sulphate into potassium and

sulphate ions is shown below as another example:

K2SO4(s) → 2K+(aq) + SO2−4 (aq)

Remember that molecular substances (e.g. covalent compounds) may also dissolve, but most will not form ions.

One example is sugar.

C6H12O6(s) C6H12O6(aq)

There are exceptions to this and some molecular substances will form ions when they dissolve. Hydrogen

chloride for example can ionise to form hydrogen and chloride ions.

HCl (g) → H+(aq) + Cl−(aq)

NOTE: The ability of ionic compounds to dissolve in water is extremely important in

the human body! The body is made up of cells, each of which is surrounded by a

membrane. Dissolved ions are found inside and outside of body cells in different con-

centrations. Some of these ions are positive (e.g. Mg2+) and some are negative (e.g.

Cl−). If there is a difference in the charge that is inside and outside the cell, then

there is a potential difference across the cell membrane. This is called the membrane

potential of the cell. The membrane potential acts like a battery and affects the move-

ment of all charged substances across the membrane. Membrane potentials play a

role in muscle functioning, digestion, excretion and in maintaining blood pH to name

just a few. The movement of ions across the membrane can also be converted into an

electric signal that can be transferred along neurons (nerve cells), which control body

processes. If ionic substances were not able to dissociate in water, then none of these

processes would be possible! It is also important to realise that our bodies can lose

ions such as Na+, K+, Ca2+, Mg2+, and Cl−, for example when we sweat during ex-

ercise. Sports drinks such as Lucozade and Powerade are designed to replace these

lost ions so that the body’s normal functioning is not affected.

Exercise 10.1: Dissociation in water Write a balanced equation to

show how silver nitrate (AgNO3) dissociates in water.

Solution to Exercise

Step 1. The cation is: Ag+ and the anion is: NO−3

Step 2. Since we know both the anion and the cation that silver nitrate dis-

sociates into we can write the following equation:

AgNO3(s) → Ag+(aq) + NO−3 (aq) (10.1)

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Ions in solution

1. For each of the following, say whether the substance is ionic or molecular.

a. potassium nitrate (KNO3)

b. ethanol (C2H5OH)

c. sucrose (a type of sugar) (C12H22O11)

d. sodium bromide (NaBr)

2. Write a balanced equation to show how each of the following ionic compounds dissociate in water.

a. sodium sulphate (Na2SO4)

b. potassium bromide (KBr)c. potassium permanganate (KMnO4)

d. sodium phosphate (Na3PO4)

Find the answers with the shortcodes:

(1.) l33 (2.) l3g

1.69.2 Ions and water hardness

This section is not examinable and is included as an example of ions in aqueous solution.

Definition: Water hardness

Water hardness is a measure of the mineral content of water. Minerals are substances

such as calcite, quartz and mica that occur naturally as a result of geological processes.

Hard water is water that has a high mineral content. Water that has a low mineral content is known as soft

water. If water has a high mineral content, it usually contains high levels of metal ions, mainly calcium (Ca) and

magnesium (Mg). The calcium enters the water from either CaCO3 (limestone or chalk) or from mineral deposits

of CaSO4. The main source of magnesium is a sedimentary rock called dolomite, CaMg (CO3)2. Hard water

may also contain other metals as well as bicarbonates and sulphates.

NOTE: The simplest way to check whether water is hard or soft is to use the lather/froth

test. If the water is very soft, soap will lather more easily when it is rubbed against the

skin. With hard water this won’t happen. Toothpaste will also not froth well in hard

water.

A water softener works on the principle of ion exchange. Hard water passes through a media bed, usually

made of resin beads that are supersaturated with sodium. As the water passes through the beads, the hardness

minerals (e.g. calcium and magnesium) attach themselves to the beads. The sodium that was originally on the

beads is released into the water. When the resin becomes saturated with calcium and magnesium, it must be

recharged. A salt solution is passed through the resin. The sodium replaces the calcium and magnesium and

these ions are released into the waste water and discharged.

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CHAPTER 1. REACTIONS IN AQUEOUS SOLUTIONS 1.70. ACID RAIN

1.70 Acid rain

(section shortcode: C10071 )

This section is not examinable and is included as an example of ions in aqueous solution.

The acidity of rainwater comes from the natural presence of three substances (CO2, NO, and SO2) in the low-

est layer of the atmosphere. These gases are able to dissolve in water and therefore make rain more acidic

than it would otherwise be. Of these gases, carbon dioxide (CO2) has the highest concentration and therefore

contributes the most to the natural acidity of rainwater. We will look at each of these gases in turn.

Definition: Acid rain

Acid rain refers to the deposition of acidic components in rain, snow and dew. Acid rain

occurs when sulphur dioxide and nitrogen oxides are emitted into the atmosphere, undergo

chemical transformations and are absorbed by water droplets in clouds. The droplets then

fall to earth as rain, snow, mist, dry dust, hail, or sleet. This increases the acidity of the soil

and affects the chemical balance of lakes and streams.

1. Carbon dioxide Carbon dioxide reacts with water in the atmosphere to form carbonic acid (H2CO3).

CO2 +H2O → H2CO3 (10.2)

The carbonic acid dissociates to form hydrogen and hydrogen carbonate ions. It is the presence of hydrogen

ions that lowers the pH of the solution making the rain acidic.

H2CO3 → H+ +HCO−3 (10.3)

2. Nitric oxide Nitric oxide (NO) also contributes to the natural acidity of rainwater and is formed during

lightning storms when nitrogen and oxygen react. In air, NO is oxidised to form nitrogen dioxide (NO2). It is

the nitrogen dioxide which then reacts with water in the atmosphere to form nitric acid (HNO3).

3NO2 (g) + H2O(l) → 2HNO3 (aq) + NO (g) (10.4)

The nitric acid dissociates in water to produce hydrogen ions and nitrate ions. This again lowers the pH of

the solution making it acidic.

HNO3 → H+ +NO−3 (10.5)

3. Sulphur dioxide Sulphur dioxide in the atmosphere first reacts with oxygen to form sulphur trioxide, before

reacting with water to form sulphuric acid.

2SO2 +O2 → 2SO3 (10.6)

SO3 +H2O → H2SO4 (10.7)

Sulphuric acid dissociates in a similar way to the previous reactions.

H2SO4 → HSO−4 +H+ (10.8)

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Although these reactions do take place naturally, human activities can greatly increase the concentration of these

gases in the atmosphere, so that rain becomes far more acidic than it would otherwise be. The burning of fossil

fuels in industries, vehicles etc is one of the biggest culprits. If the acidity of the rain drops to below 5, it is referred

to as acid rain.

Acid rain can have a very damaging effect on the environment. In rivers, dams and lakes, increased acidity

can mean that some species of animals and plants will not survive. Acid rain can also degrade soil minerals,

producing metal ions that are washed into water systems. Some of these ions may be toxic e.g. Al3+. From an

economic perspective, altered soil pH can drastically affect agricultural productivity.

Acid rain can also affect buildings and monuments, many of which are made from marble and limestone. A

chemical reaction takes place between CaCO3 (limestone) and sulphuric acid to produce aqueous ions which

can be easily washed away. The same reaction can occur in the lithosphere where limestone rocks are present

e.g. limestone caves can be eroded by acidic rainwater.

H2SO4 +CaCO3 → CaSO4 ·H2O+CO2 (10.9)

1.70.1 Investigation : Acid rain

You are going to test the effect of ’acid rain’ on a number of substances.

Materials needed:

samples of chalk, marble, zinc, iron, lead, dilute sulphuric acid, test tubes, beaker, glass dropper

Method:

1. Place a small sample of each of the following substances in a separate test tube: chalk, marble, zinc, iron

and lead

2. To each test tube, add a few drops of dilute sulphuric acid.

3. Observe what happens and record your results.

Discussion questions:

• In which of the test tubes did reactions take place? What happened to the sample substances?

• What do your results tell you about the effect that acid rain could have on each of the following: buildings,

soils, rocks and geology, water ecosystems?

• What precautions could be taken to reduce the potential impact of acid rain?

1.71 Electrolytes, ionisation and conductivity

(section shortcode: C10072 )

Conductivity in aqueous solutions, is a measure of the ability of water to conduct an electric current. The more

ions there are in the solution, the higher its conductivity.

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Definition: Conductivity

Conductivity is a measure of a solution’s ability to conduct an electric current.

1.71.1 Electrolytes

An electrolyte is a material that increases the conductivity of water when dissolved in it. Electrolytes can be

further divided into strong electrolytes and weak electrolytes.

Definition: Electrolyte

An electrolyte is a substance that contains free ions and behaves as an electrically con-

ductive medium. Because they generally consist of ions in solution, electrolytes are also

known as ionic solutions.

1. Strong electrolytes A strong electrolyte is a material that ionises completely when it is dissolved in water:

AB (s, l, g) → A+(aq) + B−(aq) (10.10)

This is a chemical change because the original compound has been split into its component ions and

bonds have been broken. In a strong electrolyte, we say that the extent of ionisation is high. In other words,

the original material dissociates completely so that there is a high concentration of ions in the solution. An

example is a solution of potassium nitrate:

KNO3(s) → K+(aq) + NO−3 (aq) (10.11)

2. Weak electrolytes A weak electrolyte is a material that goes into solution and will be surrounded by water

molecules when it is added to water. However, not all of the molecules will dissociate into ions. The extent

of ionisation of a weak electrolyte is low and therefore the concentration of ions in the solution is also low.

AB (s, l, g) → AB (aq) A+ (aq) +B− (aq) (10.12)

The following example shows that in the final solution of a weak electrolyte, some of the original compound

plus some dissolved ions are present.

C2H3O2H(l) → C2H3O2H C2H3O−2 (aq) + H+ (aq) (10.13)

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1.71.2 Non-electrolytes

A non-electrolyte is a material that does not increase the conductivity of water when dissolved in it. The sub-

stance goes into solution and becomes surrounded by water molecules, so that the molecules of the chemical

become separated from each other. However, although the substance does dissolve, it is not changed in any

way and no chemical bonds are broken. The change is a physical change. In the oxygen example below, the

reaction is shown to be reversible because oxygen is only partially soluble in water and comes out of solution

very easily.

C2H5OH(l) → C2H5OH(aq) (10.14)

O2 (g) O2 (aq) (10.15)

1.71.3 Factors that affect the conductivity of water

The conductivity of water is therefore affected by the following factors:

• The type of substance that dissolves in water. Whether a material is a strong electrolyte (e.g. potassium

nitrate, KNO3), a weak electrolyte (e.g. acetate, CH3COOH) or a non-electrolyte (e.g. sugar, alcohol, oil)

will affect the conductivity of water because the concentration of ions in solution will be different in each

case.

• The concentration of ions in solution. The higher the concentration of ions in solution, the higher its

conductivity will be.

• Temperature. The warmer the solution, the higher the solubility of the material being dissolved and there-

fore the higher the conductivity as well.

Experiment : Electrical conductivity

Aim:

To investigate the electrical conductivities of different substances and solutions.

Apparatus:

Solid salt (NaCl) crystals; different liquids such as distilled water, tap water, seawater, benzene and alcohol;

solutions of salts e.g. NaCl, KBr; a solution of an acid (e.g. HCl) and a solution of a base (e.g. NaOH); torch

cells; ammeter; conducting wire, crocodile clips and 2 carbon rods.

Method:

Set up the experiment by connecting the circuit as shown in the diagram below. In the diagram, ’X’ represents

the substance or solution that you will be testing. When you are using the solid crystals, the crocodile clips can

be attached directly to each end of the crystal. When you are using solutions, two carbon rods are placed into

the liquid and the clips are attached to each of the rods. In each case, complete the circuit and allow the current

to flow for about 30 seconds. Observe whether the ammeter shows a reading.

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Results:

Record your observations in a table similar to the one below:

Test substance Ammeter reading

Table 10.1

What do you notice? Can you explain these observations?

Remember that for electricity to flow, there needs to be a movement of charged particles e.g. ions. With the

solid NaCl crystals, there was no flow of electricity recorded on the ammeter. Although the solid is made up of

ions, they are held together very tightly within the crystal lattice and therefore no current will flow. Distilled water,

benzene and alcohol also don’t conduct a current because they are covalent compounds and therefore do not

contain ions.

The ammeter should have recorded a current when the salt solutions and the acid and base solutions were

connected in the circuit. In solution, salts dissociate into their ions, so that these are free to move in the solution.

Acids and bases behave in a similar way and dissociate to form hydronium and oxonium ions. Look at the

following examples:

KBr → K+ + Br− (10.16)

NaCl → Na+ +Cl− (10.17)

HCl + H2O → H3O+ +Cl− (10.18)

NaOH → Na+ +OH− (10.19)

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Conclusions:

Solutions that contain free-moving ions are able to conduct electricity because of the movement of charged

particles. Solutions that do not contain free-moving ions do not conduct electricity.

NOTE: Conductivity in streams and rivers is affected by the geology of the area where

the water is flowing through. Streams that run through areas with granite bedrock

tend to have lower conductivity because granite is made of materials that do not ionise

when washed into the water. On the other hand, streams that run through areas with

clay soils tend to have higher conductivity because the materials ionise when they are

washed into the water. Pollution can also affect conductivity. A failing sewage system

or an inflow of fertiliser runoff would raise the conductivity because of the presence

of chloride, phosphate, and nitrate ions, while an oil spill (non-ionic) would lower the

conductivity. It is very important that conductivity is kept within a certain acceptable

range so that the organisms living in these water systems are able to survive.

1.72 Types of reactions

(section shortcode: C10073 )

We will look at three types of reactions that occur in aqueous solutions. These are precipitation reactions, acid-

base reactions and redox reactions. Precipitation and acid-base reactions are sometimes called ion exchange

reactions. Redox reactions are electron transfer reactions. It is important to remember the difference between

these two types of reactions. In ion exchange reactions ions are exchanged, in electron transfer reactions elec-

trons are transferred. These terms will be explained further in the following sections.

Ion exchange reactions can be represented by:

AB(aq) + CD (aq) → AD+CB (10.20)

Either AD or CB may be a solid or a gas. When a solid forms this is known as a precipitation reaction. If a gas is

formed then this may be called a gas forming reaction. Acid-base reactions are a special class of ion exchange

reactions and we will look at them seperately.

The formation of a precipitate or a gas helps to make the reaction happen. We say that the reaction is driven by

the formation of a precipitate or a gas. All chemical reactions will only take place if there is something to make

them happen. For some reactions this happens easily and for others it is harder to make the reaction occur.

Definition: Ion exchange reaction

A type of reaction where the positive ions exchange their respective negative ions due to

a driving force.

NOTE: Ion exchange reactions are used in ion exchange chromatography. Ion ex-

change chromatography is used to purify water and as a means of softening water.

Often when chemists talk about ion exchange, they mean ion exchange chromatogra-

phy.

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1.72.1 Precipitation reactions

Sometimes, ions in solution may react with each other to form a new substance that is insoluble. This is called a

precipitate.

Definition: Precipitate

A precipitate is the solid that forms in a solution during a chemical reaction.

Demonstration : The reaction of ions in solution

Apparatus and materials:

4 test tubes; copper(II) chloride solution; sodium carbonate solution; sodium sulphate solution

Method:

1. Prepare 2 test tubes with approximately 5 ml of dilute Cu(II) chloride solution in each

2. Prepare 1 test tube with 5 ml sodium carbonate solution

3. Prepare 1 test tube with 5 ml sodium sulphate solution

4. Carefully pour the sodium carbonate solution into one of the test tubes containing copper(II) chloride and

observe what happens

5. Carefully pour the sodium sulphate solution into the second test tube containing copper(II) chloride and

observe what happens

Results:

1. A light blue precipitate forms when sodium carbonate reacts with copper(II) chloride

2. No precipitate forms when sodium sulphate reacts with copper(II) chloride

It is important to understand what happened in the previous demonstration. We will look at what happens in each

reaction, step by step.

1. Reaction 1: Sodium carbonate reacts with copper(II) chloride.

When these compounds react, a number of ions are present in solution: Cu2+, Cl−, Na+ and CO2−3 .

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Because there are lots of ions in solution, they will collide with each other and may recombine in different

ways. The product that forms may be insoluble, in which case a precipitate will form, or the product will be

soluble, in which case the ions will go back into solution. Let’s see how the ions in this example could have

combined with each other:

Cu2+ +CO2−3 → CuCO3 (10.21)

Cu2+ + 2Cl− → CuCl2 (10.22)

Na+ +Cl− → NaCl (10.23)

2Na+ +CO2−3 → Na2CO3 (10.24)

You can automatically exclude the reactions where sodium carbonate and copper(II) chloride are the prod-

ucts because these were the initial reactants. You also know that sodium chloride (NaCl) is soluble in water,

so the remaining product (copper carbonate) must be the one that is insoluble. It is also possible to look

up which salts are soluble and which are insoluble. If you do this, you will find that most carbonates are

insoluble, therefore the precipitate that forms in this reaction must be CuCO3. The reaction that has taken

place between the ions in solution is as follows:

2Na+ +CO2−3 +Cu2+ + 2Cl− → CuCO3 + 2Na+ + 2Cl− (10.25)

2. Reaction 2: Sodium sulphate reacts with copper(II) chloride.

The ions that are present in solution are Cu2+, Cl−, Na+ and SO2−4 . The ions collide with each other and

may recombine in different ways. The possible combinations of the ions are as follows:

Cu2+ + SO2−4 → CuSO4 (10.26)

Cu2+ + 2Cl− → CuCl2 (10.27)

Na+ +Cl− → NaCl (10.28)

Na+ + SO2−4 → Na2SO4 (10.29)

If we look up which of these salts are soluble and which are insoluble, we see that most chlorides and

most sulphates are soluble. This is why no precipitate forms in this second reaction. Even when the ions

recombine, they immediately separate and go back into solution. The reaction that has taken place between

the ions in solution is as follows:

2Na+ + SO2−4 +Cu2+ + 2Cl− → 2Na+ + SO2−

4 +Cu2+ + 2Cl− (10.30)

Table 10.2 shows some of the general rules about the solubility of different salts based on a number of investiga-

tions:

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Salt Solubility

Nitrates All are soluble

Potassium, sodium and ammonium salts All are soluble

Chlorides, bromides and iodides All are soluble except silver, lead(II) and mercury(II)

salts (e.g. silver chloride)

Sulphates All are soluble except lead(II) sulphate, barium sul-

phate and calcium sulphate

Carbonates All are insoluble except those of potassium, sodium

and ammonium

Compounds with fluorine Almost all are soluble except those of magnesium,

calcium, strontium (II), barium (II) and lead (II)

Perchlorates and acetates All are soluble

Chlorates All are soluble except potassium chlorate

Metal hydroxides and oxides Most are insoluble

Table 10.2: General rules for the solubility of salts

Salts of carbonates, phosphates, oxalates, chromates and sulphides are generally insoluble.

1.72.2 Testing for common anions in solution

It is also possible to carry out tests to determine which ions are present in a solution. You should try to do each

of these tests in class.

Test for a chloride

Prepare a solution of the unknown salt using distilled water and add a small amount of silver nitrate solution. If

a white precipitate forms, the salt is either a chloride or a carbonate.

Cl− +Ag+ +NO−3 → AgCl + NO−

3 (10.31)

(AgCl is white precipitate)

CO2−3 + 2Ag+ + 2NO−

3 → Ag2CO3 + 2NO−3 (10.32)

(Ag2CO3 is white precipitate)

The next step is to treat the precipitate with a small amount of concentrated nitric acid. If the precipitate remains

unchanged, then the salt is a chloride. If carbon dioxide is formed, and the precipitate disappears, the salt is a

carbonate.

AgCl + HNO3 → (no reaction; precipitate is unchanged)

Ag2CO3 + 2HNO3 → 2AgNO3 +H2O+CO2 (precipitate disappears)

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Test for a sulphate

Add a small amount of barium chloride solution to a solution of the test salt. If a white precipitate forms, the salt

is either a sulphate or a carbonate.

SO2−4 + Ba2+ +Cl− → BaSO4 +Cl− (BaSO4 is a white precipitate)

CO2−3 + Ba2+ +Cl− → BaCO3 +Cl− (BaCO3 is a white precipitate)

If the precipitate is treated with nitric acid, it is possible to distinguish whether the salt is a sulphate or a carbonate

(as in the test for a chloride).

BaSO4 +HNO3 → (no reaction; precipitate is unchanged)

BaCO3 + 2HNO3 → Ba(NO3)2 +H2O+CO2 (precipitate disappears)

Test for a carbonate

If a sample of the dry salt is treated with a small amount of acid, the production of carbon dioxide is a positive

test for a carbonate.

Acid + CO2−3 → CO2

If the gas is passed through limewater and the solution becomes milky, the gas is carbon dioxide.

Ca (OH)2 +CO2 → CaCO3 +H2O (It is the insoluble CaCO3 precipitate that makes the limewater go milky)

Test for bromides and iodides

As was the case with the chlorides, the bromides and iodides also form precipitates when they are reacted with

silver nitrate. Silver chloride is a white precipitate, but the silver bromide and silver iodide precipitates are both

pale yellow. To determine whether the precipitate is a bromide or an iodide, we use chlorine water and carbon

tetrachloride (CCl4).

Chlorine water frees bromine gas from the bromide and colours the carbon tetrachloride a reddish brown.

Chlorine water frees iodine gas from an iodide and colours the carbon tetrachloride purple.

Precipitation reactions and ions in solution

1. Silver nitrate (AgNO3) reacts with potassium chloride (KCl) and a white precipitate is formed.

a. Write a balanced equation for the reaction that takes place.

b. What is the name of the insoluble salt that forms?

c. Which of the salts in this reaction are soluble?

2. Barium chloride reacts with sulphuric acid to produce barium sulphate and hydrochloric acid.

a. Write a balanced equation for the reaction that takes place.

b. Does a precipitate form during the reaction?

c. Describe a test that could be used to test for the presence of barium sulphate in the products.

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3. A test tube contains a clear, colourless salt solution. A few drops of silver nitrate solution are added to the

solution and a pale yellow precipitate forms. Which one of the following salts was dissolved in the original

solution?

a. NaIb. KClc. K2CO3

d. Na2SO4

(IEB Paper 2, 2005)

Find the answers with the shortcodes:

(1.) l3c (2.) l3O (3.) l3x

1.72.3 Other reactions in aqueous solutions

There are many types of reactions that can occur in aqueous solutions. In this section we will look at two of them:

acid-base reactions and redox reactions. These reactions will be covered in more detail in Grade 11.

Acid-base reactions

Acid base reactions take place between acids and bases. In general, the products will be water and a salt (i.e.

an ionic compound). An example of this type of reaction is:

NaOH(aq) + HCl (aq) → NaCl (aq) + H2O (l) (10.33)

This is an special case of an ion exchange reaction since the sodium in the sodium hydroxide swaps places with

the hydrogen in the hydrogen chloride forming sodium chloride. At the same time the hydroxide and the hydrogen

combine to form water.

Redox reactions

Redox reactions involve the exchange of electrons. One ion loses electrons and becomes more positive, while

the other ion gains electrons and becomes more negative. To decide if a redox reaction has occurred we look at

the charge of the atoms, ions or molecules involved. If one of them has become more positive and the other one

has become more negative then a redox reaction has occurred. For example, sodium metal is oxidised to form

sodium oxide (and sometimes sodium peroxide as well). The balanced equation for this is:

4Na + O2 → 2Na2O (10.34)

In the above reaction sodium and oxygen are both neutral and so have no charge. In the products however,

the sodium atom has a charge of +1 and the oxygen atom has a charge of −2. This tells us that the sodium

has lost electrons and the oxygen has gained electrons. Since one species has become more positive and one

more negative we can conclude that a redox reaction has occurred. We could also say that electrons have been

transferred from one species to the other. (In this case the electrons were transferred from the sodium to the

oxygen).

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Demonstration: Oxidation of sodium metal

You will need a bunsen burner, a small piece of sodium metal and a metal spatula. Light the bunsen burner.

Place the sodium metal on the spatula. Place the sodium in the flame. When the reaction finishes, you should

observe a white powder on the spatula. This is a mixture of sodium oxide (Na2O) and sodium peroxide (Na2O2).

WARNING: Sodium metal is very reactive. Sodium metal reacts vigourously with water

and should never be placed in water. Be very careful when handling sodium metal.

1.72.4 Experiment: Reaction types

Aim:

To use experiments to determine what type of reaction occurs.

Apparatus:

Soluble salts (e.g. potassium nitrate, ammonium chloride, sodium carbonate, silver nitrate, sodium bromide);

hydrochloric acid (HCl); sodium hydroxide(NaOH); bromothymol blue; zinc metal; copper (II) sulphate; beakers;

test-tubes

Method:

• For each of the salts, dissolve a small amount in water and observe what happens.

• Try dissolving pairs of salts (e.g. potassium nitrate and sodium carbonate) in water and observe what

happens.

• Dissolve some sodium carbonate in hydrochloric acid and observe what happens.

• Carefully measure out 20cm3 of sodium hydroxide into a beaker.

• Add some bromothymol blue to the sodium hydroxide

• Carefully add a few drops of hydrochloric acid to the sodium hydroxide and swirl. Repeat until you notice

the colour change.

• Place the zinc metal into the copper sulphate solution and observe what happens.

Results:

Answer the following questions:

• What did you observe when you dissolved each of the salts in water?

• What did you observe when you dissolved pairs of salts in the water?

• What did you observe when you dissolved sodium carbonate in hydrochloric acid?

• Why do you think we used bromothymol blue when mixing the hydrochloric acid and the sodium hydroxide?

Think about the kind of reaction that occurred.

• What did you observe when you placed the zinc metal into the copper sulphate?

• Classify each reaction as either precipitation, gas forming, acid-base or redox.

• What makes each reaction happen (i.e. what is the driving force)? Is it the formation of a precipitate or

something else?

• What criteria would you use to determine what kind of reaction occurs?

• Try to write balanced chemical equations for each reaction

Conclusion:

We can see how we can classify reactions by performing experiments.

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CHAPTER 1. REACTIONS IN AQUEOUS SOLUTIONS 1.73. SUMMARY

In the experiment above, you should have seen how each reaction type differs from the others. For example, a

gas forming reaction leads to bubbles in the solution, a precipitation reaction leads to a precipitate forming, an

acid-base reaction can be seen by adding a suitable indicator and a redox reaction can be seen by one metal

disappearing and a deposit forming in the solution.

1.73 Summary

(section shortcode: C10074 )

• The polar nature of water means that ionic compounds dissociate easily in aqueous solution into their

component ions.

• Ions in solution play a number of roles. In the human body for example, ions help to regulate the internal

environment (e.g. controlling muscle function, regulating blood pH). Ions in solution also determine water

hardness and pH.

• Water hardness is a measure of the mineral content of water. Hard water has a high mineral concentration

and generally also a high concentration of metal ions e.g. calcium and magnesium. The opposite is true for

soft water.

• Conductivity is a measure of a solution’s ability to conduct an electric current.

• An electrolyte is a substance that contains free ions and is therefore able to conduct an electric current.

Electrolytes can be divided into strong and weak electrolytes, based on the extent to which the substance

ionises in solution.

• A non-electrolyte cannot conduct an electric current because it dooes not contain free ions.

• The type of substance, the concentration of ions and the temperature of the solution affect its conduc-

tivity.

• There are three main types of reactions that occur in aqueous solutions. These are precipitation reactions,

acid-base reactions and redox reactions.

• Precipitation and acid-base reactions are sometimes known as ion exchange reactions. Ion exchange

reactions also include gas forming reactions.

• A precipitate is formed when ions in solution react with each other to form an insoluble product. Solubility

’rules’ help to identify the precipitate that has been formed.

• A number of tests can be used to identify whether certain anions are present in a solution.

• An acid-base reaction is one in which an acid reacts with a base to form a salt and water.

• A redox reaction is one in which electrons are transferred from one substance to another.

1.74 End of chapter exercises

(section shortcode: C10075 )

1. Give one word for each of the following descriptions:

a. the change in phase of water from a gas to a liquid

b. a charged atom

c. a term used to describe the mineral content of water

d. a gas that forms sulphuric acid when it reacts with water

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2. Match the information in column A with the information in column B by writing only the letter (A to I) next to

the question number (1 to 7)

Column A Column B

1. A polar molecule A. H2SO4

2. molecular solution B. CaCO3

3. Mineral that increases water hardness C. NaOH

4. Substance that increases the hydrogen ion concentration D. salt water

5. A strong electrolyte E. calcium

6. A white precipitate F. carbon dioxide

7. A non-conductor of electricity G. potassium nitrate

H. sugar water

I. O2

Table 10.3

3. For each of the following questions, choose the one correct answer from the list provided.

a. Which one of the following substances does not conduct electricity in the solid phase but is an electrical

conductor when molten?

i. Cuii. PbBr2

iii. H2Oiv. I2

(IEB Paper 2, 2003)

b. The following substances are dissolved in water. Which one of the solutions is basic?

i. sodium nitrate

ii. calcium sulphate

iii. ammonium chloride

iv. potassium carbonate

(IEB Paper 2, 2005)

4. Explain the difference between a weak electrolyte and a strong electrolyte. Give a generalised equation for

each.

5. What factors affect the conductivity of water? How do each of these affect the conductivity?

6. For each of the following substances state whether they are molecular or ionic. If they are ionic, give a

balanced reaction for the dissociation in water.

a. Methane (CH4)

b. potassium bromide

c. carbon dioxide

d. hexane (C6H14)

e. lithium fluoride (LiF)

f. magnesium chloride

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7. Three test tubes (X, Y and Z) each contain a solution of an unknown potassium salt. The following ob-

servations were made during a practical investigation to identify the solutions in the test tubes: A: A white

precipitate formed when silver nitrate (AgNO3) was added to test tube Z. B: A white precipitate formed in test

tubes X and Y when barium chloride (BaCl2) was added. C: The precipitate in test tube X dissolved in hy-

drochloric acid (HCl) and a gas was released. D: The precipitate in test tube Y was insoluble in hydrochloric

acid.

a. Use the above information to identify the solutions in each of the test tubes X, Y and Z.

b. Write a chemical equation for the reaction that took place in test tube X before hydrochloric acid was

added.

(DoE Exemplar Paper 2 2007)

Find the answers with the shortcodes:

(1.) l3a (2.) l3C (3a.) l31 (3b.) l3r (4.) lTl (5.) lTi

(6.) lbk (7.) l3Y

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Quantitative Aspects of Chemical Change

1.75 Quantitative Aspects of Chemical Change

(section shortcode: C10076 )

An equation for a chemical reaction can provide us with a lot of useful information. It tells us what the reactants

and the products are in the reaction, and it also tells us the ratio in which the reactants combine to form products.

Look at the equation below:

Fe + S → FeS

In this reaction, every atom of iron (Fe) will react with a single atom of sulphur (S) to form one molecule of iron

sulphide (FeS). However, what the equation doesn’t tell us, is the quantities or the amount of each substance

that is involved. You may for example be given a small sample of iron for the reaction. How will you know how

many atoms of iron are in this sample? And how many atoms of sulphur will you need for the reaction to use

up all the iron you have? Is there a way of knowing what mass of iron sulphide will be produced at the end of

the reaction? These are all very important questions, especially when the reaction is an industrial one, where it

is important to know the quantities of reactants that are needed, and the quantity of product that will be formed.

This chapter will look at how to quantify the changes that take place in chemical reactions.

1.76 The Mole

(section shortcode: C10077 )

Sometimes it is important to know exactly how many particles (e.g. atoms or molecules) are in a sample of a

substance, or what quantity of a substance is needed for a chemical reaction to take place.

You will remember from Relative atomic mass (Section 3.7.2: Relative atomic mass) that the relative atomic

mass of an element, describes the mass of an atom of that element relative to the mass of an atom of carbon-12.

So the mass of an atom of carbon (relative atomic mass is 12 u) for example, is twelve times greater than the

mass of an atom of hydrogen, which has a relative atomic mass of 1 u. How can this information be used to help

us to know what mass of each element will be needed if we want to end up with the same number of atoms of

carbon and hydrogen?

Let’s say for example, that we have a sample of 12 g carbon. What mass of hydrogen will contain the same

number of atoms as 12 g carbon? We know that each atom of carbon weighs twelve times more than an atom

of hydrogen. Surely then, we will only need 1 g of hydrogen for the number of atoms in the two samples to be

the same? You will notice that the number of particles (in this case, atoms) in the two substances is the same

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when the ratio of their sample masses (12 g carbon: 1g hydrogen = 12:1) is the same as the ratio of their relative

atomic masses (12 u: 1 u = 12:1).

To take this a step further, if you were to weigh out samples of a number of elements so that the mass of the

sample was the same as the relative atomic mass of that element, you would find that the number of particles in

each sample is 6, 022×1023. These results are shown in Table 11.1 below for a number of different elements. So,

24, 31 g of magnesium (relative atomic mass = 24, 31 u) for example, has the same number of atoms as 40, 08 gof calcium (relative atomic mass = 40, 08 u).

Element Relative atomic mass (u) Sample mass (g) Atoms in sample

Hydrogen (H) 1 1 6, 022× 1023

Carbon (C) 12 12 6, 022× 1022

Magnesium (Mg) 24.31 24.31 6, 022× 1023

Sulphur (S) 32.07 32.07 6, 022× 1023

Calcium (Ca) 40.08 40.08 6, 022× 1023

Table 11.1: Table showing the relationship between the sample mass, the relative atomic mass and the number

of atoms in a sample, for a number of elements.

This result is so important that scientists decided to use a special unit of measurement to define this quantity:

the mole or ’mol’. A mole is defined as being an amount of a substance which contains the same number of

particles as there are atoms in 12 g of carbon. In the examples that were used earlier, 24, 31 g magnesium is one

mole of magnesium, while 40, 08 g of calcium is one mole of calcium. A mole of any substance always contains

the same number of particles.

Definition: Mole

The mole (abbreviation ’n’) is the SI (Standard International) unit for ’amount of substance’.

It is defined as an amount of substance that contains the same number of particles (atoms,

molecules or other particle units) as there are atoms in 12 g carbon.

In one mole of any substance, there are 6, 022× 1023 particles.

Definition: Avogadro’s number

The number of particles in a mole, equal to 6, 022× 1023. It is also sometimes referred to

as the number of atoms in 12 g of carbon-12.

If we were to write out Avogadro’s number then it would look like: 602200000000000000000000. This is a very

large number. If we had this number of cold drink cans, then we could cover the surface of the earth to a depth

of over 300 km! If you could count atoms at a rate of 10 million per second, then it would take you 2 billion years

to count the atoms in one mole!

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We can build up to the idea of Avogadro’s number. For example, if you have 12 eggs then you have a dozen

eggs. After this number we get a gross of eggs, which is 144 eggs. Finally if we wanted one mole of eggs this

would be 6, 022× 1023. That is a lot of eggs!

NOTE: The original hypothesis that was proposed by Amadeo Avogadro was that

’equal volumes of gases, at the same temperature and pressure, contain the same

number of molecules’. His ideas were not accepted by the scientific community and

it was only four years after his death, that his original hypothesis was accepted and

that it became known as ’Avogadro’s Law’. In honour of his contribution to science, the

number of particles in one mole was named Avogadro’s number.

1.76.1 Moles and mass

1. Complete the following table:

Element Relative atomic mass (u) Sample mass (g) Number of moles in the sample

Hydrogen 1.01 1.01

Magnesium 24.31 24.31

Carbon 12.01 24.02

Chlorine 35.45 70.9

Nitrogen 42.08

Table 11.2

2. How many atoms are there in...

a. 1 mole of a substance

b. 2 moles of calcium

c. 5 moles of phosphorus

d. 24, 31 g of magnesium

e. 24, 02 g of carbon

Find the answers with the shortcodes:

(1.) lgl (2.) lgi

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1.77 Molar Mass

(section shortcode: C10078 )

Definition: Molar mass

Molar mass (M) is the mass of 1 mole of a chemical substance. The unit for molar mass is

grams per mole or g ·mol−1.

Refer to Table 11.1. You will remember that when the mass, in grams, of an element is equal to its relative atomic

mass, the sample contains one mole of that element. This mass is called the molar mass of that element.

You may sometimes see the molar mass written as Mm. We will use M in this book, but you should be aware of

the alternate notation.

It is worth remembering the following: On the periodic table, the relative atomic mass that is shown can be

interpreted in two ways.

1. The mass of a single, average atom of that element relative to the mass of an atom of carbon.

2. The mass of one mole of the element. This second use is the molar mass of the element.

Element Relative atomic mass

(u)

Molar mass (g ·mol−1) Mass of one mole of

the element (g)

Magnesium 24,31 24,31 24,31

Lithium 6,94 6,94 6,94

Oxygen 16 16 16

Nitrogen 14,01 14,01 14,01

Iron 55,85 55,85 55,85

Table 11.3: The relationship between relative atomic mass, molar mass and the mass of one mole for a number

of elements.

Exercise 11.1: Calculating the number of moles from mass Calcu-

late the number of moles of iron (Fe) in a 11, 7 g sample.

Solution to Exercise

Step 1. If we look at the periodic table, we see that the molar mass of iron

is 55, 85 g ·mol−1. This means that 1 mole of iron will have a mass

of 55, 85 g.

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Step 2. If 1 mole of iron has a mass of 55, 85 g, then: the number of moles

of iron in 111, 7 g must be:

111, 7g

55, 85g ·mol−1 = 2mol (11.1)

There are 2 moles of iron in the sample.

Exercise 11.2: Calculating mass from moles You have a sample that

contains 5 moles of zinc.

1. What is the mass of the zinc in the sample?

2. How many atoms of zinc are in the sample?

Solution to Exercise

Step 1. Molar mass of zinc is 65, 38 g ·mol−1, meaning that 1 mole of zinc

has a mass of 65, 38 g.

Step 2. If 1 mole of zinc has a mass of 65, 38 g, then 5 moles of zinc has a

mass of: 65, 38 g × 5 mol = 326, 9 g (answer to a)

Step 3.

5× 6, 022× 1023 = 30, 115× 1023 (11.2)

(answer to b)

1.77.1 Moles and molar mass

1. Give the molar mass of each of the following elements:

a. hydrogen

b. nitrogen

c. bromine

2. Calculate the number of moles in each of the following samples:

a. 21, 62 g of boron (B)

b. 54, 94 g of manganese (Mn)

c. 100, 3 g of mercury (Hg)

d. 50 g of barium (Ba)

e. 40 g of lead (Pb)

Find the answers with the shortcodes:

(1.) lg3 (2.) lgO

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1.78. AN EQUATION TO CALCULATE MOLES AND MASS IN CHEMICAL REACTIONSCHAPTER 1. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE

1.78 An equation to calculate moles and mass in chemical reactions

(section shortcode: C10079 )

The calculations that have been used so far, can be made much simpler by using the following equation:

n (number of moles) =m (mass of substance in g)

M(

molar mass of substance in g ·mol−1) (11.3)

TIP: Remember that when you use the equation n = mM

, the mass is always in grams

(g) and molar mass is in grams per mol (g ·mol−1).

The equation can also be used to calculate mass and molar mass, using the following equations:

m = n×M (11.4)

and

M =m

n(11.5)

The following diagram may help to remember the relationship between these three variables. You need to

imagine that the horizontal line is like a ’division’ sign and that the vertical line is like a ’multiplication’ sign. So,

for example, if you want to calculate ’M’, then the remaining two letters in the triangle are ’m’ and ’n’ and ’m’ is

above ’n’ with a division sign between them. In your calculation then, ’m’ will be the numerator and ’n’ will be the

denominator.

Exercise 11.3: Calculating moles from mass Calculate the number

of moles of copper there are in a sample that weighs 127 g.

Solution to Exercise

Step 1.

n =m

M(11.6)

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Step 2.

n =127

63, 55= 2 (11.7)

There are 2 moles of copper in the sample.

Exercise 11.4: Calculating mass from moles You are given a 5 mol

sample of sodium. What mass of sodium is in the sample?

Solution to Exercise

Step 1.

m = n×M (11.8)

Step 2. MNa = 22, 99 g ·mol−1

Therefore,

m = 5× 22, 99 = 114, 95 g (11.9)

The sample of sodium has a mass of 114, 95 g.

Exercise 11.5: Calculating atoms from mass Calculate the number

of atoms there are in a sample of aluminium that weighs 80, 94 g.

Solution to Exercise

Step 1.

n =m

M=

80, 94

26, 98= 3moles (11.10)

Step 2. Number of atoms in 3 mol aluminium = 3× 6, 022× 1023

There are 18, 069× 1023 aluminium atoms in a sample of 80, 94 g.

1.78.1 Some simple calculations

1. Calculate the number of moles in each of the following samples:

a. 5, 6 g of calcium

b. 0, 02 g of manganese

c. 40 g of aluminium

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2. A lead sinker has a mass of 5 g.

a. Calculate the number of moles of lead the sinker contains.

b. How many lead atoms are in the sinker?

3. Calculate the mass of each of the following samples:

a. 2, 5 mol magnesium

b. 12 mol lithium

c. 4, 5× 1025 atoms of silicon

Find the answers with the shortcodes:

(1.) lgc (2.) lga (3.) lgx

1.79 Molecules and compounds

(section shortcode: C10080 )

So far, we have only discussed moles, mass and molar mass in relation to elements. But what happens if we are

dealing with a molecule or some other chemical compound? Do the same concepts and rules apply? The answer

is ’yes’. However, you need to remember that all your calculations will apply to the whole molecule. So, when

you calculate the molar mass of a molecule, you will need to add the molar mass of each atom in that compound.

Also, the number of moles will also apply to the whole molecule. For example, if you have one mole of nitric acid

(HNO3), it means you have 6, 022× 1023 molecules of nitric acid in the sample. This also means that there are

6, 022× 1023 atoms of hydrogen, 6, 022× 1023 atoms of nitrogen and (3× 6, 022× 1023) atoms of oxygen in the

sample.

In a balanced chemical equation, the number that is written in front of the element or compound, shows the mole

ratio in which the reactants combine to form a product. If there are no numbers in front of the element symbol,

this means the number is ’1’.

e.g. N2 + 3H2 → 2NH3

In this reaction, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia.

Exercise 11.6: Calculating molar mass Calculate the molar mass of

H2SO4.

Solution to Exercise

Step 1. Hydrogen = 1, 008 g ·mol−1; Sulphur = = 32, 07 g ·mol−1; Oxygen

= 16 g ·mol−1

Step 2.

M(H2SO4) = (2× 1, 008) + (32, 07) + (4× 16) = 98, 09 g ·mol−1 (11.11)

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Exercise 11.7: Calculating moles from mass Calculate the number

of moles there are in 1 kg of MgCl2.

Solution to Exercise

Step 1.

n =m

M(11.12)

Step 2. a. Convert mass into grams

m = 1kg × 1000 = 1000 g (11.13)

b. Calculate the molar mass of MgCl2.

M(MgCl2) = 24, 31 + (2× 35, 45) = 95, 21 g ·mol−1 (11.14)

Step 3.

n =1000

95, 21= 10, 5 mol (11.15)

There are 10, 5 moles of magnesium chloride in a 1 kg sample.

Exercise 11.8: Calculating the mass of reactants and products Bar-

ium chloride and sulphuric acid react according to the following equation

to produce barium sulphate and hydrochloric acid.

BaCl2 +H2SO4 → BaSO4 + 2HCl

If you have 2 g of BaCl2...

1. What quantity (in g) of H2SO4 will you need for the reaction so that

all the barium chloride is used up?

2. What mass of HCl is produced during the reaction?

Solution to Exercise

Step 1.

n =m

M=

2

208, 24= 0, 0096 mol (11.16)

Step 2. According to the balanced equation, 1 mole of BaCl2 will react with

1 mole of H2SO4. Therefore, if 0, 0096 mol of BaCl2 react, then

there must be the same number of moles of H2SO4 that react be-

cause their mole ratio is 1:1.

Step 3.

m = n×M = 0, 0096× 98, 086 = 0, 94 g (11.17)

(answer to 1)

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Step 4. According to the balanced equation, 2 moles of HCl are produced

for every 1 mole of the two reactants. Therefore the number of

moles of HCl produced is (2×0, 0096), which equals 0, 0096 moles.Step 5.

m = n×M = 0, 0192× 35, 73 = 0, 69 g (11.18)

(answer to 2)

1.79.1 Group work : Understanding moles, molecules and Avogadro’s number

Divide into groups of three and spend about 20 minutes answering the following questions together:

1. What are the units of the mole? Hint: Check the definition of the mole.

2. You have a 56 g sample of iron sulphide (FeS)

a. How many moles of FeS are there in the sample?

b. How many molecules of FeS are there in the sample?

c. What is the difference between a mole and a molecule?

3. The exact size of Avogadro’s number is sometimes difficult to imagine.

a. Write down Avogadro’s number without using scientific notation.

b. How long would it take to count to Avogadro’s number? You can assume that you can count two

numbers in each second.

Khan academy video on the mole - 1 (Video: P10081 )

1.79.2 More advanced calculations

1. Calculate the molar mass of the following chemical compounds:

a. KOHb. FeCl3c. Mg (OH)2

2. How many moles are present in:

a. 10 g of Na2SO4

b. 34 g of Ca (OH)2c. 2, 45× 1023 molecules of CH4?

3. For a sample of 0, 2 moles of potassium bromide (KBr), calculate...

a. the number of moles of K+ ions

b. the number of moles of Br− ions

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4. You have a sample containing 3 moles of calcium chloride.

a. What is the chemical formula of calcium chloride?

b. How many calcium atoms are in the sample?

5. Calculate the mass of:

a. 3 moles of NH4OHb. 4, 2 moles of Ca (NO3) 2

6. 96, 2 g sulphur reacts with an unknown quantity of zinc according to the following equation: Zn + S → ZnS

a. What mass of zinc will you need for the reaction, if all the sulphur is to be used up?

b. What mass of zinc sulphide will this reaction produce?

7. Calcium chloride reacts with carbonic acid to produce calcium carbonate and hydrochloric acid according to

the following equation: CaCl2 + H2CO3 → CaCO3 + 2HCl If you want to produce 10 g of calcium carbon-

ate through this chemical reaction, what quantity (in g) of calcium chloride will you need at the start of the

reaction?

Find the answers with the shortcodes:

(1.) lgC (2.) lgr (3.) lg1 (4.) lgY (5.) lgg (6.) lg4

(7.) lg2

1.80 The Composition of Substances

(section shortcode: C10082 )

The empirical formula of a chemical compound is a simple expression of the relative number of each type of

atom in that compound. In contrast, the molecular formula of a chemical compound gives the actual number of

atoms of each element found in a molecule of that compound.

Definition: Empirical formula

The empirical formula of a chemical compound gives the relative number of each type of

atom in that compound.

Definition: Molecular formula

The molecular formula of a chemical compound gives the exact number of atoms of each

element in one molecule of that compound.

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The compound ethanoic acid for example, has the molecular formula CH3COOH or simply C2H4O2. In one

molecule of this acid, there are two carbon atoms, four hydrogen atoms and two oxygen atoms. The ratio of

atoms in the compound is 2:4:2, which can be simplified to 1:2:1. Therefore, the empirical formula for this

compound is CH2O. The empirical formula contains the smallest whole number ratio of the elements that make

up a compound.

Knowing either the empirical or molecular formula of a compound, can help to determine its composition in more

detail. The opposite is also true. Knowing the composition of a substance can help you to determine its formula.

There are four different types of composition problems that you might come across:

1. Problems where you will be given the formula of the substance and asked to calculate the percentage by

mass of each element in the substance.

2. Problems where you will be given the percentage composition and asked to calculate the formula.

3. Problems where you will be given the products of a chemical reaction and asked to calculate the formula of

one of the reactants. These are often referred to as combustion analysis problems.

4. Problems where you will be asked to find number of moles of waters of crystallisation.

Exercise 11.9: Calculating the percentage by mass of elements in

a compound Calculate the percentage that each element contributes

to the overall mass of sulphuric acid (H2SO4).

Solution to Exercise

Step 1. Hydrogen = 1, 008× 2 = 2, 016 uSulphur = 32, 07 uOxygen = 4× 16 = 64 u

Step 2. Use the calculations in the previous step to calculate the molecular

mass of sulphuric acid.

Mass = 2, 016 + 32, 07 + 64 = 98, 09 u (11.19)

Step 3. Use the equation:

Percentage by mass = atomic massmolecular mass of H2SO4

× 100%Hydrogen

2, 016

98, 09× 100% = 2, 06% (11.20)

Sulphur

32, 07

98, 09× 100% = 32, 69% (11.21)

Oxygen

64

98, 09× 100% = 65, 25% (11.22)

(You should check at the end that these percentages add up to

100%!)

In other words, in one molecule of sulphuric acid, hydrogen makes

up 2,06% of the mass of the compound, sulphur makes up 32,69%

and oxygen makes up 65,25%.

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Exercise 11.10: Determining the empirical formula of a compound

A compound contains 52.2% carbon (C), 13.0% hydrogen (H) and 34.8%

oxygen (O). Determine its empirical formula.

Solution to Exercise

Step 1. Carbon = 52, 2 g, hydrogen = 13, 0 g and oxygen = 34, 8 gStep 2.

n =m

M(11.23)

Therefore,

n (Carbon) =52, 2

12, 01= 4, 35mol (11.24)

n (Hydrogen) =13, 0

1, 008= 12, 90mol (11.25)

n (Oxygen) =34, 8

16= 2, 18mol (11.26)

Step 3. In this case, the smallest number of moles is 2.18. Therefore...

Carbon

4, 35

2, 18= 2 (11.27)

Hydrogen

12, 90

2, 18= 6 (11.28)

Oxygen

2, 18

2, 18= 1 (11.29)

Therefore the empirical formula of this substance is: C2H6O. Do

you recognise this compound?

Exercise 11.11: Determining the formula of a compound 207 g of

lead combines with oxygen to form 239 g of a lead oxide. Use this infor-

mation to work out the formula of the lead oxide (Relative atomic masses:

Pb = 207 u and O = 16 u).

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Solution to Exercise

Step 1.

239− 207 = 32 g (11.30)

Step 2.

n =m

M(11.31)

Lead

207

207= 1mol (11.32)

Oxygen

32

16= 2mol (11.33)

Step 3. The mole ratio of Pb : O in the product is 1:2, which means that for

every atom of lead, there will be two atoms of oxygen. The formula

of the compound is PbO2.

Exercise 11.12: Empirical and molecular formula Vinegar, which is

used in our homes, is a dilute form of acetic acid. A sample of acetic acid

has the following percentage composition: 39,9% carbon, 6,7% hyrogen

and 53,4% oxygen.

1. Determine the empirical formula of acetic acid.

2. Determine the molecular formula of acetic acid if the molar mass of

acetic acid is 60 g ·mol−1.

Solution to Exercise

Step 1. In 100 g of acetic acid, there is 39, 9 g C, 6, 7 g H and 53, 4 g OStep 2. n = m

M

nC = 39,912 = 3, 33 mol

nH = 6,71 = 6, 7 mol

nO = 53,416 = 3, 34 mol

(11.34)

Step 3. Empirical formula is CH2OStep 4. The molar mass of acetic acid using the empirical formula is 30 g ·

mol−1. Therefore the actual number of moles of each element must

be double what it is in the empirical formula.

The molecular formula is therefore C2H4O2 or CH3COOH

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Exercise 11.13: Waters of crystallisation Aluminium trichloride

(AlCl3) is an ionic substance that forms crystals in the solid phase. Wa-

ter molecules may be trapped inside the crystal lattice. We represent this

as: AlCl3 · nH2O. A learner heated some aluminium trichloride crystals

until all the water had evaporated and found that the mass after heating

was 2, 8 g. The mass before heating was 5 g. What is the number of

moles of water molecules in the aluminium trichloride?

Solution to Exercise

Step 1. We first need to find n, the number of water molecules that are

present in the crystal. To do this we first note that the mass of

water lost is 5− 2, 8 = 2, 2.

Step 2. The next step is to work out the mass ratio of aluminium trichloride

to water and the mole ratio. The mass ratio is:

2, 8 : 2, 2 (11.35)

To work out the mole ratio we divide the mass ratio by the molecular

mass of each species:

2, 8

133:2, 2

18= 0, 021 : 0, 12 (11.36)

Next we do the following:

0, 0211

0, 021= 1 (11.37)

and0, 12

0, 021= 6 (11.38)

So the mole ratio of aluminium trichloride to water is:

1 : 6 (11.39)

Step 3. And now we know that there are 6 moles of water molecules in the

crystal.

Khan academy video on molecular and empirical formulae - 1 (Video: P10083 )

Khan academy video on mass composition - 1 (Video: P10084 )

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1.80.1 Moles and empirical formulae

1. Calcium chloride is produced as the product of a chemical reaction.

a. What is the formula of calcium chloride?

b. What percentage does each of the elements contribute to the mass of a molecule of calcium chloride?

c. If the sample contains 5 g of calcium chloride, what is the mass of calcium in the sample?

d. How many moles of calcium chloride are in the sample?

2. 13 g of zinc combines with 6, 4 g of sulphur. What is the empirical formula of zinc sulphide?

a. What mass of zinc sulphide will be produced?

b. What percentage does each of the elements in zinc sulphide contribute to its mass?

c. Determine the formula of zinc sulphide.

3. A calcium mineral consisted of 29,4% calcium, 23,5% sulphur and 47,1% oxygen by mass. Calculate the

empirical formula of the mineral.

4. A chlorinated hydrocarbon compound was analysed and found to consist of 24,24% carbon, 4,04% hy-

drogen and 71,72% chlorine. From another experiment the molecular mass was found to be 99g · mol−1.

Deduce the empirical and molecular formula.

Find the answers with the shortcodes:

(1.) lgT (2.) lgb (3.) lgj (4.) lgD

1.81 Molar Volumes of Gases

(section shortcode: C10085 )

It is possible to calculate the volume of one mole of gas at STP using what we know about gases.

1. Write down the ideal gas equation pV = nRT, therefore V = nRTp

2. Record the values that you know, making sure that they are in SI units You know that the gas is under

STP conditions. These are as follows: p = 101, 3 kPa = 101300 Pa n = 1mol R = 8, 31 J · K−1 ·mol−1

T = 273 K3. Substitute these values into the original equation.

V =nRT

p(11.40)

V =1mol× 8, 31J ·K−1 ·mol−1 × 273K

101300Pa(11.41)

4. Calculate the volume of 1 mole of gas under these conditions The volume of 1 mole of gas at STP is 22, 4×10−3 m3 = 22, 4dm3.

TIP: The standard units used for this equation are P in Pa, V in m3 and T in K.

Remember also that 1 000 cm3 = 1dm3 and 1 000 dm3 = 1m3.

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Exercise 11.14: Ideal Gas A sample of gas occupies a volume of

20 dm3, has a temperature of 200 K and has a pressure of 105 Pa. Cal-

culate the number of moles of gas that are present in the sample.

Solution to Exercise

Step 1. The only value that is not in SI units is volume. V = 0, 02 m3.

Step 2. We know that pV = nRTTherefore,

n =pV

RT(11.42)

Step 3.

n =105× 0, 02

8, 31× 280=

2, 1

2326, 8= 0, 0009 moles (11.43)

1.81.1 Using the combined gas law

1. An enclosed gas(i.e. one in a sealed container) has a volume of 300 cm3 and a temperature of 300 K. The

pressure of the gas is 50 kPa. Calculate the number of moles of gas that are present in the container.

2. What pressure will 3 mol of gaseous nitrogen exert if it is pumped into a container that has a volume of

25 dm3 at a temperature of 29 0 C?

3. The volume of air inside a tyre is 19 litres and the temperature is 290 K. You check the pressure of your

tyres and find that the pressure is 190 kPa. How many moles of air are present in the tyre?

4. Compressed carbon dioxide is contained within a gas cylinder at a pressure of 700 kPa. The temperature

of the gas in the cylinder is 310 K and the number of moles of gas is 13 mols of carbon dioxide. What is the

volume of the gas inside the cylinder?

Find the answers with the shortcodes:

(1.) lgW (2.) lgZ (3.) lgB (4.) lgK

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1.82. MOLAR CONCENTRATIONS OF LIQUIDSCHAPTER 1. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE

1.82 Molar concentrations of liquids

(section shortcode: C10086 )

A typical solution is made by dissolving some solid substance in a liquid. The amount of substance that is

dissolved in a given volume of liquid is known as the concentration of the liquid. Mathematically, concentration

(C) is defined as moles of solute (n) per unit volume (V) of solution.

C =n

V(11.44)

For this equation, the units for volume are dm3. Therefore, the unit of concentration is mol · dm−3. When

concentration is expressed in mol · dm−3 it is known as the molarity (M) of the solution. Molarity is the most

common expression for concentration.

TIP: Do not confuse molarity (M) with molar mass (M). Look carefully at the question

in which the M appears to determine whether it is concentration or molar mass.

Definition: Concentration

Concentration is a measure of the amount of solute that is dissolved in a given volume of

liquid. It is measured in mol ·dm−3. Another term that is used for concentration is molarity

(M)

Exercise 11.15: Concentration Calculations 1 If 3, 5 g of sodium hy-

droxide (NaOH) is dissolved in 2, 5 dm3 of water, what is the concentra-

tion of the solution in mol · dm−3?

Solution to Exercise

Step 1.

n =m

M=

3, 5

40= 0, 0875 mol (11.45)

Step 2.

C =n

V=

0, 0875

2, 5= 0, 035 (11.46)

The concentration of the solution is 0, 035 mol · dm−3 or 0, 035M

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Exercise 11.16: Concentration Calculations 2 You have a 1 dm3

container in which to prepare a solution of potassium permanganate

(KMnO4). What mass of KMnO4 is needed to make a solution with

a concentration of 0, 2M?

Solution to Exercise

Step 1.

C =n

V(11.47)

therefore

n = C × V = 0, 2× 1 = 0, 2 mol (11.48)

Step 2.

m = n×M = 0, 2× 158, 04 = 31, 61 g (11.49)

The mass of KMnO4 that is needed is 31, 61 g.

Exercise 11.17: Concentration Calculations 3 How much sodium

chloride (in g) will one need to prepare 500 cm3 of solution with a con-

centration of 0, 01M?

Solution to Exercise

Step 1.

V =500

1000= 0, 5 dm3 (11.50)

Step 2.

n = C × V = 0, 01× 0, 5 = 0, 005 mol (11.51)

Step 3.

m = n×M = 0, 005× 58, 45 = 0, 29 g (11.52)

The mass of sodium chloride needed is 0, 29 g

1.82.1 Molarity and the concentration of solutions

1. 5, 95 g of potassium bromide was dissolved in 400 cm3 of water. Calculate its molarity.

2. 100 g of sodium chloride (NaCl) is dissolved in 450 cm3 of water.

a. How many moles of NaCl are present in solution?

b. What is the volume of water (in dm3)?

c. Calculate the concentration of the solution.

d. What mass of sodium chloride would need to be added for the concentration to become 5, 7 mol·dm−3?

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1.83. STOICHIOMETRIC CALCULATIONSCHAPTER 1. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE

3. What is the molarity of the solution formed by dissolving 80 g of sodium hydroxide (NaOH) in 500 cm3 of

water?

4. What mass (g) of hydrogen chloride (HCl) is needed to make up 1000 cm3 of a solution of concentration

1 mol · dm−3?

5. How many moles of H2SO4 are there in 250 cm3 of a 0, 8M sulphuric acid solution? What mass of acid is

in this solution?

Find the answers with the shortcodes:

(1.) lgk (2.) lg0 (3.) lg8 (4.) lg9 (5.) lgX

1.83 Stoichiometric calculations

(section shortcode: C10087 )

Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is also the

numerical relationship between reactants and products. In representing chemical change showed how to write

balanced chemical equations. By knowing the ratios of substances in a reaction, it is possible to use stoichiometry

to calculate the amount of either reactants or products that are involved in the reaction. The examples shown

below will make this concept clearer.

Exercise 11.18: Stoichiometric calculation 1 What volume of oxygen

at S.T.P. is needed for the complete combustion of 2 dm3 of propane

(C3H8)? (Hint: CO2 and H2O are the products in this reaction (and in all

combustion reactions))

Solution to Exercise

Step 1. C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O(g)Step 2. From the balanced equation, the ratio of oxygen to propane in the

reactants is 5:1.

Step 3. 1 volume of propane needs 5 volumes of oxygen, therefore 2 dm3

of propane will need 10 dm3 of oxygen for the reaction to proceed

to completion.

Exercise 11.19: Stoichiometric calculation 2 What mass of iron (II)

sulphide is formed when 5, 6 g of iron is completely reacted with sulphur?

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CHAPTER 1. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE1.83. STOICHIOMETRIC CALCULATIONS

Solution to Exercise

Step 1. Fe (s) + S (s) → FeS (s)Step 2.

n =m

M=

5, 6

55, 85= 0, 1mol (11.53)

Step 3. From the equation 1 mole of Fe gives 1 mole of FeS. Therefore,

0, 1 moles of iron in the reactants will give 0, 1 moles of iron sulphide

in the product.

Step 4.

m = n×M = 0, 1× 87, 911 = 8, 79 g (11.54)

The mass of iron (II) sulphide that is produced during this reaction

is 8, 79 g.

When we are given a known mass of a reactant and are asked to work out how much product is formed, we

are working out the theoretical yield of the reaction. In the laboratory chemists never get this amount of product.

In each step of a reaction a small amount of product and reactants is ’lost’ either because a reactant did not

completely react or some of the product was left behind in the original container. Think about this. When you

make your lunch or supper, you might be a bit hungry, so you eat some of the food that you are preparing. So

instead of getting the full amount of food out (theoretical yield) that you started preparing, you lose some along

the way.

Exercise 11.20: Industrial reaction to produce fertiliser Sulphuric

acid (H2SO4) reacts with ammonia (NH3) to produce the fertiliser am-

monium sulphate ((NH4)2SO4) according to the following equation:

H2SO4 (aq) + 2NH3 (g) → (NH4)2SO4 (aq)

What is the maximum mass of ammonium sulphate that can be obtained

from 2, 0 kg of sulphuric acid?

Solution to Exercise

Step 1.

n (H2SO4) =m

M=

2000 g

98, 078 g ·mols−1 = 20, 39 mols (11.55)

Step 2. From the balanced equation, the mole ratio of H2SO4 in the reac-

tants to (NH4) 2SO4 in the product is 1:1. Therefore, 20, 39 mols of

H2SO4 of (NH4) 2SO4.

The maximum mass of ammonium sulphate that can be produced

is calculated as follows:

m = n×M = 20, 41mol× 132 g ·mol−1 = 2694 g (11.56)

The maximum amount of ammonium sulphate that can be produced

is 2, 694 kg.

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1.84. SUMMARY CHAPTER 1. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE

Khan academy video on stoichiometry - 1 (Video: P10088 )

1.83.1 Stoichiometry

1. Diborane, B2H6, was once considered for use as a rocket fuel. The combustion reaction for diborane is:

B2H6 (g)+3O2 (g) → 2HBO2 (g)+2H2O(l) If we react 2, 37 grams of diborane, how many grams of water

would we expect to produce?

2. Sodium azide is a commonly used compound in airbags. When triggered, it has the following reaction:

2NaN3 (s) → 2Na (s) + 3N2 (g) If 23, 4 grams of sodium azide is used, how many moles of nitrogen gas

would we expect to produce?

3. Photosynthesis is a chemical reaction that is vital to the existence of life on Earth. During photosynthesis,

plants and bacteria convert carbon dioxide gas, liquid water, and light into glucose (C6H12O6) and oxygen

gas.

a. Write down the equation for the photosynthesis reaction.

b. Balance the equation.

c. If 3 moles of carbon dioxide are used up in the photosynthesis reaction, what mass of glucose will be

produced?

(Presentation: P10089 )

Find the answers with the shortcodes:

(1.) lgI (2.) lg5 (3.) lgN

1.84 Summary

(section shortcode: C10090 )

• It is important to be able to quantify the changes that take place during a chemical reaction.

• The mole (n) is a SI unit that is used to describe an amount of substance that contains the same number of

particles as there are atoms in 12 g of carbon.

• The number of particles in a mole is called the Avogadro constant and its value is 6, 022 × 1023. These

particles could be atoms, molecules or other particle units, depending on the substance.

• The molar mass (M) is the mass of one mole of a substance and is measured in grams per mole or g·mol−1.

The numerical value of an element’s molar mass is the same as its relative atomic mass. For a compound,

the molar mass has the same numerical value as the molecular mass of that compound.

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CHAPTER 1. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE 1.85. END OF CHAPTER EXERCISES

• The relationship between moles (n), mass in grams (m) and molar mass (M) is defined by the following

equation:

n =m

M(11.57)

• In a balanced chemical equation, the number in front of the chemical symbols describes the mole ratio of

the reactants and products.

• The empirical formula of a compound is an expression of the relative number of each type of atom in the

compound.

• The molecular formula of a compound describes the actual number of atoms of each element in a molecule

of the compound.

• The formula of a substance can be used to calculate the percentage by mass that each element contributes

to the compound.

• The percentage composition of a substance can be used to deduce its chemical formula.

• One mole of gas occupies a volume of 22, 4 dm3.

• The concentration of a solution can be calculated using the following equation,

C =n

V(11.58)

where C is the concentration (in mol · dm−3), n is the number of moles of solute dissolved in the solution

and V is the volume of the solution (in dm−3).

• Molarity is a measure of the concentration of a solution, and its units are mol · dm−3.

• Stoichiometry is the calculation of the quantities of reactants and products in chemical reactions. It is also

the numerical relationship between reactants and products.

• The theoretical yield of a reaction is the maximum amount of product that we expect to get out of a reaction

1.85 End of chapter exercises

(section shortcode: C10091 )

1. Write only the word/term for each of the following descriptions:

a. the mass of one mole of a substance

b. the number of particles in one mole of a substance

2. Multiple choice: Choose the one correct answer from those given.

a. 5 g of magnesium chloride is formed as the product of a chemical reaction. Select the true statement

from the answers below:

a. 0.08 moles of magnesium chloride are formed in the reaction

b. the number of atoms of Cl in the product is 0, 6022× 1023

c. the number of atoms of Mg is 0,05

d. the atomic ratio of Mg atoms to Cl atoms in the product is 1:1

b. 2 moles of oxygen gas react with hydrogen. What is the mass of oxygen in the reactants?

a. 32 g

b. 0,125 g

c. 64 g

d. 0,063 g

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1.85. END OF CHAPTER EXERCISES CHAPTER 1. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE

c. In the compound potassium sulphate (K2SO4), oxygen makes up x% of the mass of the compound. x

= ...

a. 36.8

b. 9,2

c. 4

d. 18,3

d. The molarity of a 150 cm3 solution, containing 5 g of NaCl is...

a. 0, 09Mb. 5, 7 × 10−4 Mc. 0, 57Md. 0, 03M

3. Calculate the number of moles in:

a. 5 g of methane (CH4)

b. 3,4 g of hydrochloric acid

c. 6,2 g of potassium permanganate (KMnO4)

d. 4 g of neon

e. 9,6 kg of titanium tetrachloride (TiCl4)

4. Calculate the mass of:

a. 0,2 mols of potassium hydroxide (KOH)

b. 0,47 mols of nitrogen dioxide

c. 5,2 mols of helium

d. 0,05 mols of copper (II) chloride (CuCl2)

e. 31, 31× 1023 molecules of carbon monoxide (CO)

5. Calculate the percentage that each element contributes to the overall mass of:

a. Chloro-benzene (C6H5Cl)b. Lithium hydroxide (LiOH)

6. CFC’s (chlorofluorocarbons) are one of the gases that contribute to the depletion of the ozone layer. A

chemist analysed a CFC and found that it contained 58,64% chlorine, 31,43% fluorine and 9,93% carbon.

What is the empirical formula?

7. 14 g of nitrogen combines with oxygen to form 46 g of a nitrogen oxide. Use this information to work out the

formula of the oxide.

8. Iodine can exist as one of three oxides (I2O4; I2O5; I4O9). A chemist has produced one of these oxides

and wishes to know which one they have. If he started with 508 g of iodine and formed 652 g of the oxide,

which form has he produced?

9. A fluorinated hydrocarbon (a hydrocarbon is a chemical compound containing hydrogen and carbon.) was

analysed and found to contain 8,57% H, 51,05% C and 40,38% F.

a. What is its empirical formula?

b. What is the molecular formula if the molar mass is 94, 1g ·mol−1?

10. Copper sulphate crystals often include water. A chemist is trying to determine the number of moles of water

in the copper sulphate crystals. She weighs out 3 g of copper sulphate and heats this. After heating, she

finds that the mass is 1,9 g. What is the number of moles of water in the crystals? (Copper sulphate is

represented by CuSO4 · xH2O).

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CHAPTER 1. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE 1.85. END OF CHAPTER EXERCISES

11. 300 cm3 of a 0, 1 mol · dm−3 solution of sulphuric acid is added to 200 cm3 of a 0, 5 mol · dm−3 solution of

sodium hydroxide.

a. Write down a balanced equation for the reaction which takes place when these two solutions are mixed.

b. Calculate the number of moles of sulphuric acid which were added to the sodium hydroxide solution.

c. Is the number of moles of sulphuric acid enough to fully neutralise the sodium hydroxide solution?

Support your answer by showing all relevant calculations. (IEB Paper 2 2004)

12. A learner is asked to make 200 cm3 of sodium hydroxide (NaOH) solution of concentration 0, 5 mol ·dm−3.

a. Determine the mass of sodium hydroxide pellets he needs to use to do this.

b. Using an accurate balance the learner accurately measures the correct mass of the NaOH pellets. To

the pellets he now adds exactly 200 cm3 of pure water. Will his solution have the correct concentration?

Explain your answer.

c. The learner then takes 300 cm3 of a 0, 1 mol · dm−3 solution of sulphuric acid (H2SO4) and adds it to

200 cm3 of a 0, 5 mol · dm−3 solution of NaOH at 250C.

d. Write down a balanced equation for the reaction which takes place when these two solutions are mixed.

e. Calculate the number of moles of H2SO4 which were added to the NaOH solution.

f. Is the number of moles of H2SO4 calculated in the previous question enough to fully neutralise the

NaOH solution? Support your answer by showing all the relevant calculations. (IEB Paper 2, 2004)

Find the answers with the shortcodes:

(1.) lgR (2.a) lgn (2.b) lgQ (2.c) lgU (2.d) lgy (3.) lT3

(4.) lTO (5.) lTc (6.) lTx (7.) lTa (8.) lTC (9.) lT1

(10.) lTr (11.) lgP (12.) lgm

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The Hydrosphere

1.86 Introduction

(section shortcode: C10092 )

As far as we know, the Earth we live on is the only planet that is able to support life. Amongst other factors,

the Earth is just the right distance from the sun to have temperatures that are suitable for life to exist. Also, the

Earth’s atmosphere has exactly the right type of gases in the right amounts for life to survive. Our planet also has

water on its surface, which is something very unique. In fact, Earth is often called the ’Blue Planet’ because most

of it is covered in water. This water is made up of freshwater in rivers and lakes, the saltwater of the oceans and

estuaries, groundwater and water vapour. Together, all these water bodies are called the hydrosphere.

NOTE: The total mass of the hydrosphere is approximately 1, 4 × 1018 tonnes! (The

volume of one tonne of water is approximately 1 cubic metre.)

1.87 Interactions of the hydrosphere

(section shortcode: C10093 )

It is important to realise that the hydrosphere is not an isolated system, but rather interacts with other global sys-

tems, including the atmosphere, lithosphere and biosphere. These interactions are sometimes known collectively

as the water cycle.

• Atmosphere When water is heated (e.g. by energy from the sun), it evaporates and forms water vapour.

When water vapour cools again, it condenses to form liquid water which eventually returns to the surface by

precipitation e.g. rain or snow. This cycle of water moving through the atmosphere and the energy changes

that accompany it, is what drives weather patterns on earth.

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1.88. EXPLORING THE HYDROSPHERE CHAPTER 1. THE HYDROSPHERE

• Lithosphere In the lithosphere (the ocean and continental crust at the Earth’s surface), water is an impor-

tant weathering agent, which means that it helps to break rock down into rock fragments and then soil.

These fragments may then be transported by water to another place, where they are deposited. These two

processes (weathering and the transporting of fragments) are collectively called erosion. Erosion helps to

shape the earth’s surface. For example, you can see this in rivers. In the upper streams, rocks are eroded

and sediments are transported down the river and deposited on the wide flood plains lower down. On a

bigger scale, river valleys in mountains have been carved out by the action of water, and cliffs and caves on

rocky beach coastlines are also the result of weathering and erosion by water. The processes of weathering

and erosion also increase the content of dissolved minerals in the water. These dissolved minerals are

important for the plants and animals that live in the water.

• Biosphere In the biosphere, land plants absorb water through their roots and then transport this through

their vascular (transport) system to stems and leaves. This water is needed in photosynthesis, the food

production process in plants. Transpiration (evaporation of water from the leaf surface) then returns water

back to the atmosphere.

1.88 Exploring the Hydrosphere

(section shortcode: C10094 )

The large amount of water on our planet is something quite unique. In fact, about 71% of the earth is covered

by water. Of this, almost 97% is found in the oceans as saltwater, about 2.2% occurs as a solid in ice sheets,

while the remaining amount (less than 1%) is available as freshwater. So from a human perspective, despite the

vast amount of water on the planet, only a very small amount is actually available for human consumption (e.g.

drinking water). In Reactions in aqueous solutions (Chapter 9) we looked at some of the reactions that occur

in aqueous solution and saw some of the chemistry of water, in this section we are going to spend some time

exploring a part of the hydrosphere in order to start appreciating what a complex and beautiful part of the world it

is. After completing the following investigation, you should start to see just how important it is to know about the

chemistry of water.

1.88.1 Investigation : Investigating the hydrosphere

1. For this exercise, you can choose any part of the hydrosphere that you would like to explore. This may be

a rock pool, a lake, river, wetland or even just a small pond. The guidelines below will apply best to a river

investigation, but you can ask similar questions and gather similar data in other areas. When choosing your

study site, consider how accessible it is (how easy is it to get to?) and the problems you may experience

(e.g. tides, rain).

2. Your teacher will provide you with the equipment you need to collect the following data. You should have at

least one study site where you will collect data, but you might decide to have more if you want to compare

your results in different areas. This works best in a river, where you can choose sites down its length.

a. Chemical data Measure and record data such as temperature, pH, conductivity and dissolved oxygen

at each of your sites. You may not know exactly what these measurements mean right now, but it will

become clearer later.

b. Hydrological data Measure the water velocity of the river and observe how the volume of water in the

river changes as you move down its length. You can also collect a water sample in a clear bottle, hold

it to the light and see whether the water is clear or whether it has particles in it.

c. Biological data What types of animals and plants are found in or near this part of the hydrosphere? Are

they specially adapted to their environment?

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CHAPTER 1. THE HYDROSPHERE 1.89. THE IMPORTANCE OF THE HYDROSPHERE

Record your data in a table like the one shown below:

Site 1 Site 2 Site 3

Temperature

pH

Conductivity

Dissolved oxygen

Animals and plants

Table 12.1

3. Interpreting the data Once you have collected and recorded your data, think about the following questions:

• How does the data you have collected vary at different sites?

• Can you explain these differences?

• What effect do you think temperature, dissolved oxygen and pH have on animals and plants that are

living in the hydrosphere?

• Water is seldom ’pure’. It usually has lots of things dissolved (e.g. Mg2+, Ca2+ and NO−3 ions) or

suspended (e.g. soil particles, debris) in it. Where do these substances come from?

• Are there any human activities near this part of the hydrosphere? What effect could these activities

have on the hydrosphere?

1.89 The Importance of the Hydrosphere

(section shortcode: C10095 )

It is so easy sometimes to take our hydrosphere for granted and we seldom take the time to really think about the

role that this part of the planet plays in keeping us alive. Below are just some of the very important functions of

water in the hydrosphere:

• Water is a part of living cells Each cell in a living organism is made up of almost 75% water, and this allows

the cell to function normally. In fact, most of the chemical reactions that occur in life, involve substances

that are dissolved in water. Without water, cells would not be able to carry out their normal functions and life

could not exist.

• Water provides a habitat The hydrosphere provides an important place for many animals and plants to live.

Many gases (e.g. CO2, O2), nutrients e.g. nitrate (NO−3 ), nitrite (NO−

2 ) and ammonium (NH+4 ) ions, as well

as other ions (e.g. Mg2+ and Ca2+) are dissolved in water. The presence of these substances is critical for

life to exist in water.

• Regulating climate One of water’s unique characteristics is its high specific heat. This means that water

takes a long time to heat up and also a long time to cool down. This is important in helping to regulate

temperatures on earth so that they stay within a range that is acceptable for life to exist. Ocean currents

also help to disperse heat.

• Human needs Humans use water in a number of ways. Drinking water is obviously very important, but water

is also used domestically (e.g. washing and cleaning) and in industry. Water can also be used to generate

electricity through hydropower.

These are just a few of the very important functions that water plays on our planet. Many of the functions of water

relate to its chemistry and to the way in which it is able to dissolve substances in it.

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1.90. THREATS TO THE HYDROSPHERE CHAPTER 1. THE HYDROSPHERE

1.90 Threats to the Hydrosphere

(section shortcode: C10096 )

It should be clear by now that the hydrosphere plays an extremely important role in the survival of life on Earth

and that the unique properties of water allow various important chemical processes to take place which would

otherwise not be possible. Unfortunately for us however, there are a number of factors that threaten our hydro-

sphere and most of these threats are because of human activities. We are going to focus on two of these issues:

overuse and pollution and look at ways in which these problems can possibly be overcome.

1. Pollution

Pollution of the hydrosphere is also a major problem. When we think of pollution, we sometimes only think

of things like plastic, bottles, oil and so on. But any chemical that is present in the hydrosphere in an amount

that is not what it should be is a pollutant. Animals and plants that live in the Earth’s water bodies are

specially adapted to surviving within a certain range of conditions. If these conditions are changed (e.g.

through pollution), these organisms may not be able to survive. Pollution then, can affect entire aquatic

ecosystems. The most common forms of pollution in the hydrosphere are waste products from humans and

from industries, nutrient pollution e.g. fertiliser runoff which causes eutrophication (an excess of nutrients

in the water leading to excessive plant growth) and toxic trace elements such as aluminium, mercury and

copper to name a few. Most of these elements come from mines or from industries.

2. Overuse of water

We mentioned earlier that only a very small percentage of the hydrosphere’s water is available as freshwater.

However, despite this, humans continue to use more and more water to the point where water consumption

is fast approaching the amount of water that is available. The situation is a serious one, particularly in

countries such as South Africa which are naturally dry and where water resources are limited. It is estimated

that between 2020 and 2040, water supplies in South Africa will no longer be able to meet the growing

demand for water in this country. This is partly due to population growth, but also because of the increasing

needs of industries as they expand and develop. For each of us, this should be a very scary thought. Try to

imagine a day without water... difficult isn’t it? Water is so much a part of our lives, that we are hardly aware

of the huge part that it plays in our daily lives.

1.90.1 Discussion : Creative water conservation

As populations grow, so do the demands that are placed on dwindling water resources. While many people argue

that building dams helps to solve this water-shortage problem, there is evidence that dams are only a temporary

solution and that they often end up doing far more ecological damage than good. The only sustainable solution is

to reduce the demand for water, so that water supplies are sufficient to meet this. The more important question

then is how to do this.

Discussion:

Divide the class into groups, so that there are about five people in each. Each group is going to represent a

different sector within society. Your teacher will tell you which sector you belong to from the following: Farming,

industry, city management or civil society (i.e. you will represent the ordinary ’man on the street’). In your groups,

discuss the following questions as they relate to the group of people you represent: (Remember to take notes

during your discussions, and nominate a spokesperson to give feedback to the rest of the class on behalf of your

group)

• What steps could be taken by your group to conserve water?

• Why do you think these steps are not being taken?

• What incentives do you think could be introduced to encourage this group to conserve water more efficiently?

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CHAPTER 1. THE HYDROSPHERE 1.91. HOW PURE IS OUR WATER?

1.90.2 Investigation: Building of dams

In the previous discussion, we mentioned that there is evidence that dams are only a temporary solution to the

water crisis. In this investigation you will look at why dams are a potentially bad solution to the problem.

For this investigation you will choose a dam that has been built in your area, or an area close to you. Make a note

of which rivers are in the area. Try to answer the following questions:

• If possible talk to people who have lived in the area for a long time and try to get their opinion on how life

changed since the dam was built. If it is not possible to talk to people in the area, then look for relevant

literature on the area.

• Try to find out if any environmental impact assessments (this is where people study the environmnent and

see what effect the proposed project has on the environment) were done before the dam was built. Why do

you think this is important? Why do you think companies do not do these assessments?

• Look at how the ecology has changed. What was the ecology of the river? What is the current ecology? Do

you think it has changed in a good way or a bad way?

Write a report or give a presentation in class on your findings from this investigation. Critically examine your

findings and draw your own conclusion as to whether or not dams are only a short term solution to the growing

water crisis.

It is important to realise that our hydrosphere exists in a delicate balance with other systems and that disturbing

this balance can have serious consequences for life on this planet.

1.90.3 Group Project : School Action Project

There is a lot that can be done within a school to save water. As a class, discuss what actions could be taken by

your class to make people more aware of how important it is to conserve water. Also consider what ways your

school can save water. If possible, try to put some of these ideas into action and see if they really do conserve

water.

1.91 How pure is our water?

(section shortcode: C10097 )

When you drink a glass of water you are not just drinking water, but many other substances that are dissolved

into the water. Some of these come from the process of making the water safe for humans to drink, while others

come from the environment. Even if you took water from a mountain stream (which is often considered pure and

bottled for people to consume), the water would still have impurities in it. Water pollution increases the amount of

impurities in the water and sometimes makes the water unsafe for drinking. In this section we will look at a few of

the substances that make water impure and how we can make pure water. We will also look at the pH of water.

In Reactions in aqueous solutions (Chapter 9) we saw how compounds can dissolve in water. Most of these

compounds (e.g. Na+, Cl−, Ca2+, Mg2+, etc.) are safe for humans to consume in the small amounts that are

naturally present in water. It is only when the amounts of these ions rise above the safe levels that the water is

considered to be polluted.

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1.91. HOW PURE IS OUR WATER? CHAPTER 1. THE HYDROSPHERE

You may have noticed sometimes that when you pour a glass a water straight from the tap, it has a sharp smell.

This smell is the same smell that you notice around swimming pools and is due to chlorine in the water. Chlorine

is the most common compound added to water to make it safe for humans to use. Chlorine helps to remove

bacteria and other biological contaminants in the water. Other methods to purify water include filtration (passing

the water through a very fine mesh) and flocculation (a process of adding chemicals to the water to help remove

small particles).

pH of water is also important. Water that is to basic (pH greater than 7) or to acidic (pH less than 7) may present

problems when humans consume the water. If you have ever noticed after swimming that your eyes are red or

your skin is itchy, then the pH of the swimming pool was probably to basic or to acidic. This shows you just

how sensitive we are to the smallest changes in our environment. The pH of water depends on what ions are

dissolved in the water. Adding chlorine to water often lowers the pH. You will learn more about pH in grade 11.

1.91.1 Experiment: Water purity

Aim:

To test the purity and pH of water samples

Apparatus:

pH test strips (you can find these at pet shops, they are used to test pH of fish tanks), microscope (or magnifying

glass), filter paper, funnel, silver nitrate, concentrated nitric acid, barium chloride, acid, chlorine water (a solution

of chlorine in water), carbon tetrachloride, some test-tubes or beakers, water samples from different sources (e.g.

a river, a dam, the sea, tap water, etc.).

Method:

1. Look at each water sample and note if the water is clear or cloudy.

2. Examine each water sample under a microscope and note what you see.

3. Test the pH of each of the water samples.

4. Pour some of the water from each sample through filter paper.

5. Refer to Testing for common anions in solutions (Section 9.5.2: Testing for common anions in solution) for

the details of common anion tests. Test for chloride, sulphate, carbonate, bromide and iodide in each of the

water samples.

Results:

Write down what you saw when you just looked at the water samples. Write down what you saw when you looked

at the water samples under a microscope. Where there any dissolved particles? Or other things in the water?

Was there a difference in what you saw with just looking and with looking with a a microscope? Write down the

pH of each water sample. Look at the filter paper from each sample. Is there sand or other particles on it? Which

anions did you find in each sample?

Discussion:

Write a report on what you observed. Draw some conclusions on the purity of the water and how you can tell if

water is pure or not.

Conclusion:

You should have seen that water is not pure, but rather has many substances dissolved in it.

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CHAPTER 1. THE HYDROSPHERE 1.92. SUMMARY

1.91.2 Project: water purification

Prepare a presentation on how water is purified. This can take the form of a poster, or a presentation or a project.

Things that you should look at are:

• Water for drinking (potable water)

• Distilled water and its uses

• Deionised water and its uses

• What methods are used to prepare water for various uses

• What regulations govern drinking water

• Why water needs to be purified

• How safe are the purification methods

1.92 Summary

(section shortcode: C10098 )

• The hydrosphere includes all the water that is on Earth. Sources of water include freshwater (e.g. rivers,

lakes), saltwater (e.g. oceans), groundwater (e.g. boreholes) and water vapour. Ice (e.g. glaciers) is also

part of the hydrosphere.

• The hydrosphere interacts with other global systems, including the atmosphere, lithosphere and biosphere.

• The hydrosphere has a number of important functions. Water is a part of all living cells, it provides a habitat

for many living organisms, it helps to regulate climate and it is used by humans for domestic, industrial and

other use.

• Despite the importance of the hydrosphere, a number of factors threaten it. These include overuse of water,

and pollution.

• Water is not pure, but has many substances dissolved in it.

1.93 End of chapter exercises

(section shortcode: C10099 )

1. What is the hydrosphere? How does it interact with other global systems?

2. Why is the hydrosphere important?

Find the answers with the shortcodes:

(1.) lgp (2.) lgd

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1.93. END OF CHAPTER EXERCISES CHAPTER 1. THE HYDROSPHERE

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Part II

Physics

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Vectors

1.94 Introduction

(section shortcode: P10000 )

This chapter focuses on vectors. We will learn what a vector is and how it differs from everyday numbers. We will

also learn how to add, subtract and multiply them and where they appear in Physics.

Are vectors Physics? No, vectors themselves are not Physics. Physics is just a description of the world around

us. To describe something we need to use a language. The most common language used to describe Physics is

Mathematics. Vectors form a very important part of the mathematical description of Physics, so much so that it is

absolutely essential to master the use of vectors.

1.95 Scalars and Vectors

(section shortcode: P10001 )

In Mathematics, you learned that a number is something that represents a quantity. For example if you have 5

books, 6 apples and 1 bicycle, the 5, 6, and 1 represent how many of each item you have.

These kinds of numbers are known as scalars.

Definition: Scalar

A scalar is a quantity that has only magnitude (size).

An extension to a scalar is a vector, which is a scalar with a direction. For example, if you travel 1 km down Main

Road to school, the quantity 1 km down Main Road is a vector. The “1 km” is the quantity (or scalar) and the

“down Main Road” gives a direction.

In Physics we use the word magnitude to refer to the scalar part of the vector.

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1.96. NOTATION CHAPTER 1. VECTORS

Definition: Vectors

A vector is a quantity that has both magnitude and direction.

A vector should tell you how much and which way.

For example, a man is driving his car east along a freeway at 100 km·hr−1. What we have given here is a vector

– the velocity. The car is moving at 100 k · h−1 (this is the magnitude) and we know where it is going – east (this

is the direction). Thus, we know the speed and direction of the car. These two quantities, a magnitude and a

direction, form a vector we call velocity.

1.96 Notation

(section shortcode: P10002 )

Vectors are different to scalars and therefore have their own notation.

1.96.1 Mathematical Representation

There are many ways of writing the symbol for a vector. Vectors are denoted by symbols with an arrow pointing

to the right above it. For example,→a ,

→v and

F represent the vectors acceleration, velocity and force, meaning

they have both a magnitude and a direction.

Sometimes just the magnitude of a vector is needed. In this case, the arrow is omitted. In other words, F denotes

the magnitude of the vector→

F . |→

F | is another way of representing the magnitude of a vector.

1.96.2 Graphical Representation

Vectors are drawn as arrows. An arrow has both a magnitude (how long it is) and a direction (the direction in

which it points). The starting point of a vector is known as the tail and the end point is known as the head.

Figure 13.1: Examples of vectors

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CHAPTER 1. VECTORS 1.97. DIRECTIONS

Figure 13.2: Parts of a vector

1.97 Directions

(section shortcode: P10003 )

There are many acceptable methods of writing vectors. As long as the vector has a magnitude and a direction, it

is most likely acceptable. These different methods come from the different methods of expressing a direction for

a vector.

1.97.1 Relative Directions

The simplest method of expressing direction is with relative directions: to the left, to the right, forward, backward,

up and down.

1.97.2 Compass Directions

Another common method of expressing directions is to use the points of a compass: North, South, East, and

West. If a vector does not point exactly in one of the compass directions, then we use an angle. For example,

we can have a vector pointing 40 North of West. Start with the vector pointing along the West direction: Then

rotate the vector towards the north until there is a 40 angle between the vector and the West. The direction of

this vector can also be described as: W 40 N (West 40 North); or N 50 W (North 50 West)

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1.97. DIRECTIONS CHAPTER 1. VECTORS

1.97.3 Bearing

The final method of expressing direction is to use a bearing. A bearing is a direction relative to a fixed point.

Given just an angle, the convention is to define the angle with respect to the North. So, a vector with a direction of

110 has been rotated clockwise 110 relative to the North. A bearing is always written as a three digit number,

for example 275 or 080 (for 80).

Scalars and Vectors

1. Classify the following quantities as scalars or vectors:

a. 12 km

b. 1 m south

c. 2 m · s−1, 45

d. 075, 2 cm

e. 100 k · h−1, 0

2. Use two different notations to write down the direction of the vector in each of the following diagrams:

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CHAPTER 1. VECTORS 1.98. DRAWING VECTORS

a.

b.

c.

Find the answers with the shortcodes:

(1.) l4s (2.) l4H

1.98 Drawing Vectors

(section shortcode: P10004 )

In order to draw a vector accurately we must specify a scale and include a reference direction in the diagram. A

scale allows us to translate the length of the arrow into the vector’s magnitude. For instance if one chose a scale

of 1 cm = 2 N (1 cm represents 2 N), a force of 20 N towards the East would be represented as an arrow 10 cm

long. A reference direction may be a line representing a horizontal surface or the points of a compass.

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1.98. DRAWING VECTORS CHAPTER 1. VECTORS

Method: Drawing Vectors

1. Decide upon a scale and write it down.

2. Determine the length of the arrow representing the vector, by using the scale.

3. Draw the vector as an arrow. Make sure that you fill in the arrow head.

4. Fill in the magnitude of the vector.

Exercise 13.1: Drawing vectors Represent the following vector quan-

tities:

1. 6 m · s−1 north

2. 16 m east

Solution to Exercise

Step 1. a. 1 cm = 2 m · s−1

b. 1 cm = 4 m

Step 2. a. If 1 cm = 2 m · s−1, then 6 m · s−1 = 3 cm

b. If 1 cm = 4 m, then 16 m = 4 cm

Step 3. a. Scale used: 1 cm = 2 m · s−1 Direction = North

b. Scale used: 1 cm = 4 m Direction = East

1.98.1 Drawing Vectors

Draw each of the following vectors to scale. Indicate the scale that you have used:

1. 12 km south

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2. 1,5 m N 45 W

3. 1 m·s−1, 20 East of North

4. 50 km·hr−1, 085

5. 5 mm, 225

Find the answers with the shortcodes:

(1.) l46

1.99 Mathematical Properties of Vectors

(section shortcode: P10005 )

Vectors are mathematical objects and we need to understand the mathematical properties of vectors, like adding

and subtracting.

For all the examples in this section, we will use displacement as our vector quantity. Displacement was discussed

in Grade 10.

Displacement is defined as the distance together with direction of the straight line joining a final point to an initial

point.

Remember that displacement is just one example of a vector. We could just as well have decided to use forces

or velocities to illustrate the properties of vectors.

1.99.1 Adding Vectors

When vectors are added, we need to add both a magnitude and a direction. For example, take 2 steps in the

forward direction, stop and then take another 3 steps in the forward direction. The first 2 steps is a displacement

vector and the second 3 steps is also a displacement vector. If we did not stop after the first 2 steps, we would

have taken 5 steps in the forward direction in total. Therefore, if we add the displacement vectors for 2 steps and

3 steps, we should get a total of 5 steps in the forward direction. Graphically, this can be seen by first following the

first vector two steps forward and then following the second one three steps forward (ie. in the same direction):

We add the second vector at the end of the first vector, since this is where we now are after the first vector has

acted. The vector from the tail of the first vector (the starting point) to the head of the last (the end point) is then

the sum of the vectors. This is the head-to-tail method of vector addition.

As you can convince yourself, the order in which you add vectors does not matter. In the example above, if you

decided to first go 3 steps forward and then another 2 steps forward, the end result would still be 5 steps forward.

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The final answer when adding vectors is called the resultant. The resultant displacement in this case will be 5

steps forward.

Definition: Resultant of Vectors

The resultant of a number of vectors is the single vector whose effect is the same as the

individual vectors acting together.

In other words, the individual vectors can be replaced by the resultant – the overall effect is the same. If vectors→a and

b have a resultant→

R, this can be represented mathematically as,

R =→a +

b . (13.1)

Let us consider some more examples of vector addition using displacements. The arrows tell you how far to

move and in what direction. Arrows to the right correspond to steps forward, while arrows to the left correspond

to steps backward. Look at all of the examples below and check them.

This example says 1 step forward and then another step forward is the same as an arrow twice as long – two

steps forward.

This examples says 1 step backward and then another step backward is the same as an arrow twice as long –

two steps backward.

It is sometimes possible that you end up back where you started. In this case the net result of what you have

done is that you have gone nowhere (your start and end points are at the same place). In this case, your resultant

displacement is a vector with length zero units. We use the symbol→0 to denote such a vector:

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Check the following examples in the same way. Arrows up the page can be seen as steps left and arrows down

the page as steps right.

Try a couple to convince yourself!

Table 13.1

Table 13.2

It is important to realise that the directions are not special– ‘forward and backwards’ or ‘left and right’ are treated

in the same way. The same is true of any set of parallel directions:

Table 13.3

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1.99. MATHEMATICAL PROPERTIES OF VECTORS CHAPTER 1. VECTORS

Table 13.4

In the above examples the separate displacements were parallel to one another. However the same head-to-tail

technique of vector addition can be applied to vectors in any direction.

Table 13.5

Now you have discovered one use for vectors; describing resultant displacement – how far and in what direction

you have travelled after a series of movements.

Although vector addition here has been demonstrated with displacements, all vectors behave in exactly the same

way. Thus, if given a number of forces acting on a body you can use the same method to determine the resultant

force acting on the body. We will return to vector addition in more detail later.

1.99.2 Subtracting Vectors

What does it mean to subtract a vector? Well this is really simple; if we have 5 apples and we subtract 3 apples,

we have only 2 apples left. Now lets work in steps; if we take 5 steps forward and then subtract 3 steps forward

we are left with only two steps forward:

What have we done? You originally took 5 steps forward but then you took 3 steps back. That backward displace-

ment would be represented by an arrow pointing to the left (backwards) with length 3. The net result of adding

these two vectors is 2 steps forward:

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Thus, subtracting a vector from another is the same as adding a vector in the opposite direction (i.e. subtracting

3 steps forwards is the same as adding 3 steps backwards).

TIP: Subtracting a vector from another is the same as adding a vector in the opposite

direction.

In the problem, motion in the forward direction has been represented by an arrow to the right. Arrows to the right

are positive and arrows to the left are negative. More generally, vectors in opposite directions differ in sign (i.e.

if we define up as positive, then vectors acting down are negative). Thus, changing the sign of a vector simply

reverses its direction:

Table 13.6

Table 13.7

Table 13.8

In mathematical form, subtracting→a from

b gives a new vector→c :

→c =

b − →a

=→

b +(

− →a) (13.2)

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This clearly shows that subtracting vector→a from

b is the same as adding(

− →a)

to→

b . Look at the following

examples of vector subtraction.

1.99.3 Scalar Multiplication

What happens when you multiply a vector by a scalar (an ordinary number)?

Going back to normal multiplication we know that 2 × 2 is just 2 groups of 2 added together to give 4. We can

adopt a similar approach to understand how vector multiplication works.

1.100 Techniques of Vector Addition

(section shortcode: P10006 )

Now that you have learned about the mathematical properties of vectors, we return to vector addition in more

detail. There are a number of techniques of vector addition. These techniques fall into two main categories -

graphical and algebraic techniques.

1.100.1 Graphical Techniques

Graphical techniques involve drawing accurate scale diagrams to denote individual vectors and their resultants.

We next discuss the two primary graphical techniques, the head-to-tail technique and the parallelogram method.

The Head-to-Tail Method

In describing the mathematical properties of vectors we used displacements and the head-to-tail graphical

method of vector addition as an illustration. The head-to-tail method of graphically adding vectors is a standard

method that must be understood.

Method: Head-to-Tail Method of Vector Addition

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1. Draw a rough sketch of the situation.

2. Choose a scale and include a reference direction.

3. Choose any of the vectors and draw it as an arrow in the correct direction and of the correct length –

remember to put an arrowhead on the end to denote its direction.

4. Take the next vector and draw it as an arrow starting from the arrowhead of the first vector in the correct

direction and of the correct length.

5. Continue until you have drawn each vector – each time starting from the head of the previous vector. In this

way, the vectors to be added are drawn one after the other head-to-tail.

6. The resultant is then the vector drawn from the tail of the first vector to the head of the last. Its magnitude

can be determined from the length of its arrow using the scale. Its direction too can be determined from the

scale diagram.

Exercise 13.2: Head-to-Tail Addition I A ship leaves harbour H and

sails 6 km north to port A. From here the ship travels 12 km east to port B,

before sailing 5,5 km south-west to port C. Determine the ship’s resultant

displacement using the head-to-tail technique of vector addition.

Solution to Exercise

Step 1. Its easy to understand the problem if we first draw a quick sketch.

The rough sketch should include all of the information given in the

problem. All of the magnitudes of the displacements are shown and

a compass has been included as a reference direction. In a rough

sketch one is interested in the approximate shape of the vector di-

agram.

Step 2. The choice of scale depends on the actual question – you should

choose a scale such that your vector diagram fits the page.

It is clear from the rough sketch that choosing a scale where 1 cm

represents 2 km (scale: 1 cm = 2 km) would be a good choice in

this problem. The diagram will then take up a good fraction of an

A4 page. We now start the accurate construction.

Step 3. Starting at the harbour H we draw the first vector 3 cm long in the

direction north.

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Step 4. Since the ship is now at port A we draw the second vector 6 cm

long starting from point A in the direction east.

Step 5. Since the ship is now at port B we draw the third vector 2,25 cm long

starting from this point in the direction south-west. A protractor is

required to measure the angle of 45.

Step 6. As a final step we draw the resultant displacement from the starting

point (the harbour H) to the end point (port C). We use a ruler to

measure the length of this arrow and a protractor to determine its

direction.

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Step 7. We now use the scale to convert the length of the resultant in the

scale diagram to the actual displacement in the problem. Since we

have chosen a scale of 1 cm = 2 km in this problem the resultant

has a magnitude of 9,2 km. The direction can be specified in terms

of the angle measured either as 072,3 east of north or on a bearing

of 072,3.

Step 8. The resultant displacement of the ship is 9,2 km on a bearing of

072,3.

Exercise 13.3: Head-to-Tail Graphical Addition II A man walks 40 m

East, then 30 m North.

1. What was the total distance he walked?

2. What is his resultant displacement?

Solution to Exercise

Step 1.

Step 2. In the first part of his journey he traveled 40 m and in the second

part he traveled 30 m. This gives us a total distance traveled of 40

m + 30 m = 70 m.

Step 3. The man’s resultant displacement is the vector from where he

started to where he ended. It is the vector sum of his two sepa-

rate displacements. We will use the head-to-tail method of accurate

construction to find this vector.

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Step 4. A scale of 1 cm represents 10 m (1 cm = 10 m) is a good choice

here. Now we can begin the process of construction.

Step 5. We draw the first displacement as an arrow 4 cm long in an east-

wards direction.

Step 6. Starting from the head of the first vector we draw the second vector

as an arrow 3 cm long in a northerly direction.

Step 7. Now we connect the starting point to the end point and measure

the length and direction of this arrow (the resultant).

Step 8. To find the direction you measure the angle between the resultant

and the 40 m vector. You should get about 37.

Step 9. Finally we use the scale to convert the length of the resultant in

the scale diagram to the actual magnitude of the resultant displace-

ment. According to the chosen scale 1 cm = 10 m. Therefore 5

cm represents 50 m. The resultant displacement is then 50 m 37

north of east.

The Parallelogram Method

The parallelogram method is another graphical technique of finding the resultant of two vectors.

Method: The Parallelogram Method

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1. Make a rough sketch of the vector diagram.

2. Choose a scale and a reference direction.

3. Choose either of the vectors to be added and draw it as an arrow of the correct length in the correct direction.

4. Draw the second vector as an arrow of the correct length in the correct direction from the tail of the first

vector.

5. Complete the parallelogram formed by these two vectors.

6. The resultant is then the diagonal of the parallelogram. The magnitude can be determined from the length

of its arrow using the scale. The direction too can be determined from the scale diagram.

Exercise 13.4: Parallelogram Method of Vector Addition I A force of

F1 = 5N is applied to a block in a horizontal direction. A second force

F2 = 4N is applied to the object at an angle of 30 above the horizontal.

Determine the resultant force acting on the block using the parallelogram

method of accurate construction.

Solution to Exercise

Step 1.

Step 2. In this problem a scale of 1 cm = 1 N would be appropriate, since

then the vector diagram would take up a reasonable fraction of the

page. We can now begin the accurate scale diagram.

Step 3. Let us draw F1 first. According to the scale it has length 5 cm.

Step 4. Next we draw F2. According to the scale it has length 4 cm. We

make use of a protractor to draw this vector at 30 to the horizontal.

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Step 5. Next we complete the parallelogram and draw the diagonal.

The resultant has a measured length of 8,7 cm.

Step 6. We use a protractor to measure the angle between the horizontal

and the resultant. We get 13,3.

Step 7. Finally we use the scale to convert the measured length into the

actual magnitude. Since 1 cm = 1 N, 8,7 cm represents 8,7 N.

Therefore the resultant force is 8,7 N at 13,3 above the horizontal.

The parallelogram method is restricted to the addition of just two vectors. However, it is arguably the most intuitive

way of adding two forces acting on a point.

1.100.2 Algebraic Addition and Subtraction of Vectors

Vectors in a Straight Line

Whenever you are faced with adding vectors acting in a straight line (i.e. some directed left and some right, or

some acting up and others down) you can use a very simple algebraic technique:

Method: Addition/Subtraction of Vectors in a Straight Line

1. Choose a positive direction. As an example, for situations involving displacements in the directions west

and east, you might choose west as your positive direction. In that case, displacements east are negative.

2. Next simply add (or subtract) the magnitude of the vectors using the appropriate signs.

3. As a final step the direction of the resultant should be included in words (positive answers are in the positive

direction, while negative resultants are in the negative direction).

Let us consider a few examples.

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Exercise 13.5: Adding vectors algebraically I A tennis ball is rolled

towards a wall which is 10 m away from the ball. If after striking the

wall the ball rolls a further 2,5 m along the ground away from the wall,

calculate algebraically the ball’s resultant displacement.

Solution to Exercise

Step 1.

Step 2. We know that the resultant displacement of the ball (→xR) is equal

to the sum of the ball’s separate displacements (→x1 and

→x2):

→xR =

→x1 +

→x2 (13.3)

Since the motion of the ball is in a straight line (i.e. the ball moves

towards and away from the wall), we can use the method of alge-

braic addition just explained.

Step 3. Let’s choose the positive direction to be towards the wall. This

means that the negative direction is away from the wall.

Step 4. With right positive:

→x1 = +10, 0m · s−1

→x2 = −2, 5m · s−1

(13.4)

Step 5. Next we simply add the two displacements to give the resultant:

→xR =

(

+10m · s−1)

+(

−2, 5m · s−1)

= (+7, 5)m · s−1(13.5)

Step 6. Finally, in this case towards the wall is the positive direction, so:→xR

= 7,5 m towards the wall.

Exercise 13.6: Subtracting vectors algebraically I Suppose that a

tennis ball is thrown horizontally towards a wall at an initial velocity of 3

m·s−1to the right. After striking the wall, the ball returns to the thrower at

2 m·s−1. Determine the change in velocity of the ball.

Solution to Exercise

Step 1. A quick sketch will help us understand the problem.

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Step 2. Remember that velocity is a vector. The change in the velocity of

the ball is equal to the difference between the ball’s initial and final

velocities:

∆→v=

→v f − →

v i (13.6)

Since the ball moves along a straight line (i.e. left and right), we can

use the algebraic technique of vector subtraction just discussed.

Step 3. Choose the positive direction to be towards the wall. This means

that the negative direction is away from the wall.

Step 4.→v i = +3m · s−1

→v f = −2m · s−1

(13.7)

Step 5. Thus, the change in velocity of the ball is:

∆→v =

(

−2m · s−1)

−(

+3m · s−1)

= (−5)m · s−1(13.8)

Step 6. Remember that in this case towards the wall means a positive

velocity, so away from the wall means a negative velocity: ∆→v=

5m · s−1 away from the wall.

Resultant Vectors

1. Harold walks to school by walking 600 m Northeast and then 500 m N 40 W. Determine his resultant

displacement by using accurate scale drawings.

2. A dove flies from her nest, looking for food for her chick. She flies at a velocity of 2 m·s−1 on a bearing

of 135 and then at a velocity of 1,2 m·s−1 on a bearing of 230. Calculate her resultant velocity by using

accurate scale drawings.

3. A squash ball is dropped to the floor with an initial velocity of 2,5 m·s−1. It rebounds (comes back up) with

a velocity of 0,5 m·s−1.

a. What is the change in velocity of the squash ball?

b. What is the resultant velocity of the squash ball?

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Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting

along a straight line. When vectors are not in a straight line, i.e. at an angle to each other, the following method

can be used:

Find the answers with the shortcodes:

(1.) l2b (2.) l2j (3.) l4F

A More General Algebraic technique

Simple geometric and trigonometric techniques can be used to find resultant vectors.

Exercise 13.7: An Algebraic Solution I A man walks 40 m East, then

30 m North. Calculate the man’s resultant displacement.

Solution to Exercise

Step 1. As before, the rough sketch looks as follows:

Step 2. Note that the triangle formed by his separate displacement vectors

and his resultant displacement vector is a right-angle triangle. We

can thus use the Theorem of Pythagoras to determine the length

of the resultant. Let xR represent the length of the resultant vector.

Then:

x2R =

(

40m · s−1)2

+(

30m · s−1)2

x2R = 2 500m · s−12

xR = 50m · s−1

(13.9)

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Step 3. Now we have the length of the resultant displacement vector but not

yet its direction. To determine its direction we calculate the angle

α between the resultant displacement vector and East, by using

simple trigonometry:

tanα = oppositesideadjacentside

tanα = 3040

α = tan−1 (0, 75)

α = 36, 9

(13.10)

Step 4. The resultant displacement is then 50 m at 36,9 North of East.

This is exactly the same answer we arrived at after drawing a scale

diagram!

In the previous example we were able to use simple trigonometry to calculate the resultant displacement. This

was possible since the directions of motion were perpendicular (north and east). Algebraic techniques, however,

are not limited to cases where the vectors to be combined are along the same straight line or at right angles to

one another. The following example illustrates this.

Exercise 13.8: An Algebraic Solution II A man walks from point A to

point B which is 12 km away on a bearing of 45. From point B the man

walks a further 8 km east to point C. Calculate the resultant displace-

ment.

Solution to Exercise

Step 1.

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B^A F = 45 since the man walks initially on a bearing of 45.

Then, A^B G = B

^A F = 45 (parallel lines, alternate angles).

Both of these angles are included in the rough sketch.

Step 2. The resultant is the vector AC. Since we know both the lengths of

AB and BC and the included angle A^B C, we can use the cosine

rule:

AC2 = AB2 +BC2 − 2 ·AB ·BCcos

(

A^B C

)

= (12)2+ (8)

2 − 2 · (12) (8) cos (135)= 343, 8

AC = 18, 5km

(13.11)

Step 3. Next we use the sine rule to determine the angle θ:

sinθ8 = sin135

18,5

sinθ = 8×sin135

18,5

θ = sin−1 (0, 3058)

θ = 17, 8

(13.12)

To find F^A C, we add 45. Thus, F

^A C = 62, 8.

Step 4. The resultant displacement is therefore 18,5 km on a bearing of

062,8.

More Resultant Vectors

1. A frog is trying to cross a river. It swims at 3 m·s−1in a northerly direction towards the opposite bank. The

water is flowing in a westerly direction at 5 m·s−1. Find the frog’s resultant velocity by using appropriate

calculations. Include a rough sketch of the situation in your answer.

2. Sandra walks to the shop by walking 500 m Northwest and then 400 m N 30 E. Determine her resultant

displacement by doing appropriate calculations.

Find the answers with the shortcodes:

(1.) l4L (2.) l4M

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1.101. COMPONENTS OF VECTORS CHAPTER 1. VECTORS

1.101 Components of Vectors

(section shortcode: P10007 )

In the discussion of vector addition we saw that a number of vectors acting together can be combined to give a

single vector (the resultant). In much the same way a single vector can be broken down into a number of vectors

which when added give that original vector. These vectors which sum to the original are called components of

the original vector. The process of breaking a vector into its components is called resolving into components.

While summing a given set of vectors gives just one answer (the resultant), a single vector can be resolved into

infinitely many sets of components. In the diagrams below the same black vector is resolved into different pairs

of components. These components are shown as dashed lines. When added together the dashed vectors give

the original black vector (i.e. the original vector is the resultant of its components).

In practice it is most useful to resolve a vector into components which are at right angles to one another, usually

horizontal and vertical.

Any vector can be resolved into a horizontal and a vertical component. If→

A is a vector, then the horizontal

component of→

A is→

Ax and the vertical component is→

Ay.

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Exercise 13.9: Resolving a vector into components A motorist un-

dergoes a displacement of 250 km in a direction 30 north of east. Re-

solve this displacement into components in the directions north (→xN ) and

east (→xE).

Solution to Exercise

Step 1.

Step 2. Next we resolve the displacement into its components north and

east. Since these directions are perpendicular to one another, the

components form a right-angled triangle with the original displace-

ment as its hypotenuse.

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Notice how the two components acting together give the original

vector as their resultant.

Step 3. Now we can use trigonometry to calculate the magnitudes of the

components of the original displacement:

xN = (250) (sin30)

= 125 km(13.13)

and

xE = (250) (cos30)

= 216, 5 km(13.14)

Remember xN and xE are the magnitudes of the components –

they are in the directions north and east respectively.

1.101.1 Block on an incline

As a further example of components let us consider a block of mass m placed on a frictionless surface inclined

at some angle θ to the horizontal. The block will obviously slide down the incline, but what causes this motion?

The forces acting on the block are its weight mg and the normal force N exerted by the surface on the object.

These two forces are shown in the diagram below.

Now the object’s weight can be resolved into components parallel and perpendicular to the inclined surface.

These components are shown as dashed arrows in the diagram above and are at right angles to each other.

The components have been drawn acting from the same point. Applying the parallelogram method, the two

components of the block’s weight sum to the weight vector.

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To find the components in terms of the weight we can use trigonometry:

Fg‖ = mgsinθ

Fg⊥ = mgcosθ(13.15)

The component of the weight perpendicular to the slope Fg⊥ exactly balances the normal force N exerted by

the surface. The parallel component, however, Fg‖ is unbalanced and causes the block to slide down the slope.

1.101.2 Worked example

Exercise 13.10: Block on an incline plane Determine the force

needed to keep a 10 kg block from sliding down a frictionless slope. The

slope makes an angle of 30 with the horizontal.

Solution to Exercise

Step 1.

The force that will keep the block from sliding is equal to the parallel

component of the weight, but its direction is up the slope.

Step 2.

Fg‖ = mgsinθ

= (10) (9, 8) (sin30)

= 49N

(13.16)

Step 3. The force is 49 N up the slope.

1.101.3 Vector addition using components

Components can also be used to find the resultant of vectors. This technique can be applied to both graphical

and algebraic methods of finding the resultant. The method is simple: make a rough sketch of the problem, find

the horizontal and vertical components of each vector, find the sum of all horizontal components and the sum of

all the vertical components and then use them to find the resultant.

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Consider the two vectors,→

A and→

B, in Figure 13.64, together with their resultant,→

R.

Figure 13.64: An example of two vectors being added to give a resultant

Each vector in Figure 13.64 can be broken down into one component in the x-direction (horizontal) and one in

the y-direction (vertical). These components are two vectors which when added give you the original vector as

the resultant. This is shown in Figure 13.65 where we can see that:

A =→

Ax +→

Ay→

B =→

Bx +→

By→

R =→

Rx +→

Ry

(13.17)

But,→

Rx =→

Ax +→

Bx

and→

Ry =→

Ay +→

By

(13.18)

In summary, addition of the x components of the two original vectors gives the x component of the resultant.

The same applies to the y components. So if we just added all the components together we would get the same

answer! This is another important property of vectors.

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Figure 13.65: Adding vectors using components.

Exercise 13.11: Adding Vectors Using Components If in Fig-

ure 13.65,→

A= 5, 385m · s−1 at an angle of 21.8 to the horizontal and→

B= 5m · s−1 at an angle of 53,13 to the horizontal, find→

R.

Solution to Exercise

Step 1. The first thing we must realise is that the order that we add the vec-

tors does not matter. Therefore, we can work through the vectors

to be added in any order.

Step 2. We find the components of→

A by using known trigonometric ratios.

First we find the magnitude of the vertical component, Ay:

sinθ =Ay

A

sin21, 8 =Ay

5,385

Ay = (5, 385) (sin21, 8)

= 2m · s−1

(13.19)

Secondly we find the magnitude of the horizontal component, Ax:

cosθ = Ax

A

cos21.8 = Ax

5,385

Ax = (5, 385) (cos21, 8)

= 5m · s−1

(13.20)

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The components give the sides of the right angle triangle, for which

the original vector,→

A, is the hypotenuse.

Step 3. We find the components of→

B by using known trigonometric ratios.

First we find the magnitude of the vertical component, By:

sinθ =By

B

sin53, 13 =By

5

By = (5) (sin53, 13)

= 4m · s−1

(13.21)

Secondly we find the magnitude of the horizontal component, Bx:

cosθ = Bx

B

cos21, 8 = Bx

5,385

Bx = (5, 385) (cos53, 13)

= 5m · s−1

(13.22)

Step 4. Now we have all the components. If we add all the horizontal com-

ponents then we will have the x-component of the resultant vector,→

Rx. Similarly, we add all the vertical components then we will have

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the y-component of the resultant vector,→

Ry.

Rx = Ax +Bx

= 5m · s−1 + 3m · s−1

= 8m · s−1

(13.23)

Therefore,→

Rx is 8 m to the right.

Ry = Ay +By

= 2m · s−1 + 4m · s−1

= 6m · s−1

(13.24)

Therefore,→

Ry is 6 m up.

Step 5. Now that we have the components of the resultant, we can use the

Theorem of Pythagoras to determine the magnitude of the resul-

tant, R.

R2 = (Rx)2+ (Ry)

2

R2 = (6)2+ (8)

2

R2 = 100

∴ R = 10m · s−1

(13.25)

The magnitude of the resultant, R is 10 m. So all we have to do is

calculate its direction. We can specify the direction as the angle the

vectors makes with a known direction. To do this you only need to

visualise the vector as starting at the origin of a coordinate system.

We have drawn this explicitly below and the angle we will calculate

is labeled α.

Using our known trigonometric ratios we can calculate the value of

α;

tanα = 6m·s−1

8m·s−1

α = tan−1 6m·s−1

8m·s−1

α = 36, 8.

(13.26)

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Step 6.→

R is 10 m at an angle of 36, 8 to the positive x-axis.

Adding and Subtracting Components of Vectors

1. Harold walks to school by walking 600 m Northeast and then 500 m N 40o W. Determine his resultant

displacement by means of addition of components of vectors.

2. A dove flies from her nest, looking for food for her chick. She flies at a velocity of 2 m·s−1 on a bearing of

135o in a wind with a velocity of 1,2 m·s−1 on a bearing of 230o. Calculate her resultant velocity by adding

the horizontal and vertical components of vectors.

Find the answers with the shortcodes:

(1.) l4e (2.) l2Z

Vector Multiplication

Vectors are special, they are more than just numbers. This means that multiplying vectors is not necessarily the

same as just multiplying their magnitudes. There are two different types of multiplication defined for vectors. You

can find the dot product of two vectors or the cross product.

The dot product is most similar to regular multiplication between scalars. To take the dot product of two vectors,

you just multiply their magnitudes to get out a scalar answer. The mathematical definition of the dot product is:

→a •

b= | →a | · |→

b |cosθ (13.27)

Take two vectors→a and

b :

You can draw in the component of→

b that is parallel to→a :

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In this way we can arrive at the definition of the dot product. You find how much of→

b is lined up with→a by finding

the component of→

b parallel to→a . Then multiply the magnitude of that component, |

b |cosθ, with the magnitude

of→a to get a scalar.

The second type of multiplication, the cross product, is more subtle and uses the directions of the vectors in

a more complicated way. The cross product of two vectors,→a and

b , is written→a ×

b and the result of this

operation on→a and

b is another vector. The magnitude of the cross product of these two vectors is:

| →a ×→

b | = | →a ||→

b |sinθ (13.28)

We still need to find the direction of→a ×

b . We do this by applying the right hand rule.

Method: Right Hand Rule

1. Using your right hand:

2. Point your index finger in the direction of→a .

3. Point the middle finger in the direction of→

b .

4. Your thumb will show the direction of→a ×

b .

1.101.4 Summary

1. A scalar is a physical quantity with magnitude only.

2. A vector is a physical quantity with magnitude and direction.

3. Vectors may be represented as arrows where the length of the arrow indicates the magnitude and the

arrowhead indicates the direction of the vector.

4. The direction of a vector can be indicated by referring to another vector or a fixed point (eg. 30 from the

river bank); using a compass (eg. N 30 W); or bearing (eg. 053).

5. Vectors can be added using the head-to-tail method, the parallelogram method or the component method.

6. The resultant of a number of vectors is the single vector whose effect is the same as the individual vectors

acting together.

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1.101.5 End of chapter exercises: Vectors

1. An object is suspended by means of a light string. The sketch shows a horizontal force F which pulls the

object from the vertical position until it reaches an equilibrium position as shown. Which one of the following

vector diagrams best represents all the forces acting on the object?

A B C D

Table 13.9

2. A load of weight W is suspended from two strings. F1 and F2 are the forces exerted by the strings on

the load in the directions show in the figure above. Which one of the following equations is valid for this

situation?

a. W = F 21 + F 2

2

b. F1sin50 = F2sin30

c. F1cos50 = F2cos30

d. W = F1 + F2

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3. Two spring balances P and Q are connected by means of a piece of string to a wall as shown. A horizontal

force of 100 N is exerted on spring balance Q. What will be the readings on spring balances P and Q?

P Q

A 100 N 0 N

B 25 N 75 N

C 50 N 50 N

D 100 N 100 N

Table 13.10

4. A point is acted on by two forces in equilibrium. The forces

a. have equal magnitudes and directions.

b. have equal magnitudes but opposite directions.

c. act perpendicular to each other.

d. act in the same direction.

5. A point in equilibrium is acted on by three forces. Force F1 has components 15 N due south and 13 N due

west. What are the components of force F2?

a. 13 N due north and 20 due west

b. 13 N due north and 13 N due west

c. 15 N due north and 7 N due west

d. 15 N due north and 13 N due east

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6. Which of the following contains two vectors and a scalar?

a. distance, acceleration, speed

b. displacement, velocity, acceleration

c. distance, mass, speed

d. displacement, speed, velocity

7. Two vectors act on the same point. What should the angle between them be so that a maximum resultant is

obtained?

a. 0

b. 90

c. 180

d. cannot tell

8. Two forces, 4 N and 11 N, act on a point. Which one of the following cannot be the magnitude of a resultant?

a. 4 N

b. 7 N

c. 11 N

d. 15 N

Find the answers with the shortcodes:

(1.) l2B (2.) l2K (3.) l2k (4.) l20 (5.) l28 (6.) l29

(7.) l2X (8.) l2I

1.101.6 End of chapter exercises: Vectors - Long questions

1. A helicopter flies due east with an air speed of 150 km.h−1. It flies through an air current which moves at

200 km.h−1 north. Given this information, answer the following questions:

a. In which direction does the helicopter fly?

b. What is the ground speed of the helicopter?

c. Calculate the ground distance covered in 40 minutes by the helicopter.

2. A plane must fly 70 km due north. A cross wind is blowing to the west at 30 km.h−1. In which direction must

the pilot steer if the plane flies at a speed of 200 km.h−1 in windless conditions?

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3. A stream that is 280 m wide flows along its banks with a velocity of 1.80m.s−1. A raft can travel at a speed

of 2.50 m.s−1 across the stream. Answer the following questions:

a. What is the shortest time in which the raft can cross the stream?

b. How far does the raft drift downstream in that time?

c. In what direction must the raft be steered against the current so that it crosses the stream perpendicular

to its banks?

d. How long does it take to cross the stream in part c?

4. A helicopter is flying from place X to place Y . Y is 1000 km away in a direction 50 east of north and the

pilot wishes to reach it in two hours. There is a wind of speed 150 km.h−1 blowing from the northwest. Find,

by accurate construction and measurement (with a scale of 1 cm = 50 km.h−1), the

a. the direction in which the helicopter must fly, and

b. the magnitude of the velocity required for it to reach its destination on time.

5. An aeroplane is flying towards a destination 300 km due south from its present position. There is a wind

blowing from the north east at 120 km.h−1. The aeroplane needs to reach its destination in 30 minutes.

Find, by accurate construction and measurement (with a scale of 1 cm = 30 km.s−1), or otherwise,

a. the direction in which the aeroplane must fly and

b. the speed which the aeroplane must maintain in order to reach the destination on time.

c. Confirm your answers in the previous 2 subquestions with calculations.

6. An object of weight W is supported by two cables attached to the ceiling and wall as shown. The tensions

in the two cables are T1 and T2 respectively. Tension T1 = 1200 N. Determine the tension T2 and weight Wof the object by accurate construction and measurement or by calculation.

7. In a map-work exercise, hikers are required to walk from a tree marked A on the map to another tree marked

B which lies 2,0 km due East of A. The hikers then walk in a straight line to a waterfall in position C which

has components measured from B of 1,0 km E and 4,0 km N.

a. Distinguish between quantities that are described as being vector and scalar.

b. Draw a labelled displacement-vector diagram (not necessarily to scale) of the hikers’ complete journey.

c. What is the total distance walked by the hikers from their starting point at A to the waterfall C?

d. What are the magnitude and bearing, to the nearest degree, of the displacement of the hikers from their

starting point to the waterfall?

8. An object X is supported by two strings, A and B, attached to the ceiling as shown in the sketch. Each of

these strings can withstand a maximum force of 700 N. The weight of X is increased gradually.

a. Draw a rough sketch of the triangle of forces, and use it to explain which string will break first.

b. Determine the maximum weight of X which can be supported.

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9. A rope is tied at two points which are 70 cm apart from each other, on the same horizontal line. The total

length of rope is 1 m, and the maximum tension it can withstand in any part is 1000 N. Find the largest

mass (m), in kg, that can be carried at the midpoint of the rope, without breaking the rope. Include a vector

diagram in your answer.

Find the answers with the shortcodes:

(1.) l25 (2.) l2N (3.) l2R (4.) l2n (5.) l2E (6.) l2m

(7.) l2y (8.) l2V (9.) l2p

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Mechanical Energy

1.102 Introduction

(section shortcode: P10008 )

All objects have energy. The word energy comes from the Greek word energeia ([U+03AD]ν[U+03AD]ργǫια),

meaning activity or operation. Energy is closely linked to mass and cannot be created or destroyed. In this chapter

we will consider potential and kinetic energy.

1.103 Potential Energy

(section shortcode: P10009 )

The potential energy of an object is generally defined as the energy an object has because of its position relative

to other objects that it interacts with. There are different kinds of potential energy such as gravitional potential

energy, chemical potential energy, electrical potential energy, to name a few. In this section we will be looking at

gravitational potential energy.

Definition: Potential energy

Potential energy is the energy an object has due to its position or state.

Gravitational potential energy is the energy of an object due to its position above the surface of the Earth. The

symbol PE is used to refer to gravitational potential energy. You will often find that the words potential energy

are used where gravitational potential energy is meant. We can define potential energy (or gravitational potential

energy, if you like) as:

EP = mgh (14.1)

where EP = potential energy measured in joules (J)

m = mass of the object (measured in kg)

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g = gravitational acceleration (9, 8 m · s−2)

h = perpendicular height from the reference point (measured in m)

TIP: You may sometimes see potential energy written as PE. We will not use this

notation in this book, but you may see it in other books.

A suitcase, with a mass of 1 kg, is placed at the top of a 2 m high cupboard. By lifting the suitcase against the

force of gravity, we give the suitcase potential energy. This potential energy can be calculated using (2.1).

If the suitcase falls off the cupboard, it will lose its potential energy. Halfway down the cupboard, the suitcase will

have lost half its potential energy and will have only 9, 8 J left. At the bottom of the cupboard the suitcase will

have lost all it’s potential energy and it’s potential energy will be equal to zero.

Objects have maximum potential energy at a maximum height and will lose their potential energy as they fall.

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CHAPTER 1. MECHANICAL ENERGY 1.103. POTENTIAL ENERGY

Exercise 14.1: Gravitational potential energy A brick with a mass of

1 kg is lifted to the top of a 4m high roof. It slips off the roof and falls to

the ground. Calculate the potential energy of the brick at the top of the

roof and on the ground once it has fallen.

Solution to Exercise

Step 1. • The mass of the brick is m = 1 kg• The height lifted is h = 4m

All quantities are in SI units.

Step 2. • We are asked to find the gain in potential energy of the brick

as it is lifted onto the roof.

• We also need to calculate the potential energy once the brick

is on the ground again.

Step 3. Since the block is being lifted we are dealing with gravitational po-

tential energy. To work out EP , we need to know the mass of the

object and the height lifted. As both of these are given, we just

substitute them into the equation for EP .

Step 4.

EP = mgh

= (1) (9, 8) (4)

= 39, 2 J

(14.2)

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1.104. KINETIC ENERGY CHAPTER 1. MECHANICAL ENERGY

1.103.1 Gravitational Potential Energy

1. Describe the relationship between an object’s gravitational potential energy and its:

a. mass and

b. height above a reference point.

2. A boy, of mass 30 kg, climbs onto the roof of a garage. The roof is 2, 5 m from the ground. He now jumps

off the roof and lands on the ground.

a. How much potential energy has the boy gained by climbing on the roof?

b. The boy now jumps down. What is the potential energy of the boy when he is 1m from the ground?

c. What is the potential energy of the boy when he lands on the ground?

3. A hiker walks up a mountain, 800 m above sea level, to spend the night at the top in the first overnight hut.

The second day he walks to the second overnight hut, 500 m above sea level. The third day he returns to

his starting point, 200 m above sea level.

a. What is the potential energy of the hiker at the first hut (relative to sea level)?

b. How much potential energy has the hiker lost during the second day?

c. How much potential energy did the hiker have when he started his journey (relative to sea level)?

d. How much potential energy did the hiker have at the end of his journey?

Find the answers with the shortcodes:

(1.) lxE (2.) lxm (3.) lxy

1.104 Kinetic Energy

(section shortcode: P10010 )

Definition: Kinetic Energy

Kinetic energy is the energy an object has due to its motion.

Kinetic energy is the energy an object has because of its motion. This means that any moving object has kinetic

energy. The faster it moves, the more kinetic energy it has. Kinetic energy (EK ) is therefore dependent on the

velocity of the object. The mass of the object also plays a role. A truck of 2000 kg, moving at 100 km · hr−1, will

have more kinetic energy than a car of 500 kg, also moving at 100 km · hr−1. Kinetic energy is defined as:

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NOTE: You may sometimes see kinetic energy written as KE. This is simply another

way to write kinetic energy. We will not use this form in this book, but you may see it

written like this in other books.

EK =1

2mv2 (14.3)

Consider the 1 kg suitcase on the cupboard that was discussed earlier. When the suitcase falls, it will gain

velocity (fall faster), until it reaches the ground with a maximum velocity. The suitcase will not have any kinetic

energy when it is on top of the cupboard because it is not moving. Once it starts to fall it will gain kinetic energy,

because it gains velocity. Its kinetic energy will increase until it is a maximum when the suitcase reaches the

ground.

Exercise 14.2: Calculation of Kinetic Energy A 1 kg brick falls off a

4 m high roof. It reaches the ground with a velocity of 8, 85 m · s−1. What

is the kinetic energy of the brick when it starts to fall and when it reaches

the ground?

Solution to Exercise

Step 1. • The mass of the rock m = 1 kg• The velocity of the rock at the bottom vbottom = 8, 85 m·s−1

These are both in the correct units so we do not have to worry about

unit conversions.

Step 2. We are asked to find the kinetic energy of the brick at the top and

the bottom. From the definition we know that to work out EK , we

need to know the mass and the velocity of the object and we are

given both of these values.

Step 3. Since the brick is not moving at the top, its kinetic energy is zero.

Step 4.

EK = 12mv2

= 12 (1 kg)

(

8, 85 m · s−1)2

= 39, 2 J

(14.4)

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1.104. KINETIC ENERGY CHAPTER 1. MECHANICAL ENERGY

1.104.1 Checking units

According to the equation for kinetic energy, the unit should be kg ·m2 · s−2. We can prove that this unit is equal

to the joule, the unit for energy.

(kg)(

m · s−1)2

=(

kg ·m · s−2)

·m= N · m

(

because Force (N) = mass (kg)× acceleration(

m · s−2))

= J (Work (J) = Force (N)× distance (m))

(14.5)

We can do the same to prove that the unit for potential energy is equal to the joule:

(kg)(

m · s−2)

(m) = N · m= J

(14.6)

Exercise 14.3: Mixing Units & Energy Calculations A bullet, having

a mass of 150 g, is shot with a muzzle velocity of 960 m · s−1. Calculate

its kinetic energy.

Solution to Exercise

Step 1. • We are given the mass of the bullet m = 150 g. This is not the

unit we want mass to be in. We need to convert to kg.

Mass in grams÷ 1000 = Mass in kg

150 g ÷ 1000 = 0, 150 kg(14.7)

• We are given the initial velocity with which the bullet leaves the

barrel, called the muzzle velocity, and it is v = 960m · s−1.

Step 2. • We are asked to find the kinetic energy.

Step 3. We just substitute the mass and velocity (which are known) into the

equation for kinetic energy:

EK = 12mv2

= 12 (0, 150) (960)

2

= 69120 J

(14.8)

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CHAPTER 1. MECHANICAL ENERGY 1.105. MECHANICAL ENERGY

Kinetic Energy

1. Describe the relationship between an object’s kinetic energy and its:

a. mass and

b. velocity

2. A stone with a mass of 100 g is thrown up into the air. It has an initial velocity of 3 m · s−1. Calculate its

kinetic energy

a. as it leaves the thrower’s hand.

b. when it reaches its turning point.

3. A car with a mass of 700 kg is travelling at a constant velocity of 100 km · hr−1. Calculate the kinetic energy

of the car.

Find the answers with the shortcodes:

(1.) lad (2.) law (3.) lav

1.105 Mechanical Energy

(section shortcode: P10011 )

TIP: Mechanical energy is the sum of the gravitational potential energy and the kinetic

energy.

Mechanical energy, EM , is simply the sum of gravitational potential energy (EP ) and the kinetic energy (EK ).

Mechanical energy is defined as:

EM = EP + EK (14.9)

EM = EP + EK

EM = mgh+ 12mv2

(14.10)

TIP: You may see mechanical energy written as U . We will not use this notation in this

book, but you should be aware that this notation is sometimes used.

1.105.1 Conservation of Mechanical Energy

The Law of Conservation of Energy states:

Energy cannot be created or destroyed, but is merely changed from one form into another.

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Definition: Conservation of Energy

The Law of Conservation of Energy: Energy cannot be created or destroyed, but is merely

changed from one form into another.

So far we have looked at two types of energy: gravitational potential energy and kinetic energy. The sum of the

gravitational potential energy and kinetic energy is called the mechanical energy. In a closed system, one where

there are no external forces acting, the mechanical energy will remain constant. In other words, it will not change

(become more or less). This is called the Law of Conservation of Mechanical Energy and it states:

The total amount of mechanical energy in a closed system remains constant.

Definition: Conservation of Mechanical Energy

Law of Conservation of Mechanical Energy: The total amount of mechanical energy in a

closed system remains constant.

This means that potential energy can become kinetic energy, or vice versa, but energy cannot ’disappear’. The

mechanical energy of an object moving in the Earth’s gravitational field (or accelerating as a result of gravity) is

constant or conserved, unless external forces, like air resistance, acts on the object.

We can now use the conservation of mechanical energy to calculate the velocity of a body in freefall and show

that the velocity is independent of mass.

Show by using the law of conservation of energy that the velocity of a body in free fall is independent of its mass.

TIP: In problems involving the use of conservation of energy, the path taken by the ob-

ject can be ignored. The only important quantities are the object’s velocity (which gives

its kinetic energy) and height above the reference point (which gives its gravitational

potential energy).

TIP: In the absence of friction, mechanical energy is conserved and

EM before = EM after (14.11)

In the presence of friction, mechanical energy is not conserved. The mechanical

energy lost is equal to the work done against friction.

∆EM = EM before − EM after = work done against friction (14.12)

In general, mechanical energy is conserved in the absence of external forces. Examples of external forces are:

applied forces, frictional forces and air resistance.

In the presence of internal forces like the force due to gravity or the force in a spring, mechanical energy is

conserved.

The following simulation covers the law of conservation of energy.

(Simulation: lTd)

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CHAPTER 1. MECHANICAL ENERGY 1.105. MECHANICAL ENERGY

1.105.2 Using the Law of Conservation of Energy

Mechanical energy is conserved (in the absence of friction). Therefore we can say that the sum of the EP and the

EK anywhere during the motion must be equal to the sum of the EP and the EK anywhere else in the motion.

We can now apply this to the example of the suitcase on the cupboard. Consider the mechanical energy of the

suitcase at the top and at the bottom. We can say:

EM top = EM bottom

EP top + EK top = EP bottom + EK bottom

mgh+ 12mv2 = mgh+ 1

2mv2

(1) (9, 8) (2) + 0 = 0 + 12 (1)

(

v2)

19, 6 J = 12v

2

39, 2 = v2

v = 6, 26m · s−1

(14.13)

The suitcase will strike the ground with a velocity of 6, 26 m · s−1.

From this we see that when an object is lifted, like the suitcase in our example, it gains potential energy. As it

falls back to the ground, it will lose this potential energy, but gain kinetic energy. We know that energy cannot be

created or destroyed, but only changed from one form into another. In our example, the potential energy that the

suitcase loses is changed to kinetic energy.

The suitcase will have maximum potential energy at the top of the cupboard and maximum kinetic energy at the

bottom of the cupboard. Halfway down it will have half kinetic energy and half potential energy. As it moves down,

the potential energy will be converted (changed) into kinetic energy until all the potential energy is gone and only

kinetic energy is left. The 19, 6 J of potential energy at the top will become 19, 6 J of kinetic energy at the bottom.

Exercise 14.4: Using the Law of Conservation of Mechanical En-

ergy During a flood a tree truck of mass 100 kg falls down a waterfall.

The waterfall is 5 m high. If air resistance is ignored, calculate

1. the potential energy of the tree trunk at the top of the waterfall.

2. the kinetic energy of the tree trunk at the bottom of the waterfall.

3. the magnitude of the velocity of the tree trunk at the bottom of the

waterfall.

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1.105. MECHANICAL ENERGY CHAPTER 1. MECHANICAL ENERGY

Solution to Exercise

Step 1. • The mass of the tree trunk m = 100 kg• The height of the waterfall h = 5m. These are all in SI units so

we do not have to convert.

Step 2. • Potential energy at the top

• Kinetic energy at the bottom

• Velocity at the bottom

Step 3.

EP = mgh

EP = (100) (9, 8) (5)

EP = 4900 J

(14.14)

Step 4. The kinetic energy of the tree trunk at the bottom of the waterfall

is equal to the potential energy it had at the top of the waterfall.

Therefore KE = 4900 J.

Step 5. To calculate the velocity of the tree trunk we need to use the equa-

tion for kinetic energy.

EK = 12mv2

4900 = 12 (100)

(

v2)

98 = v2

v = 9, 899...

v = 9, 90m · s−1downwards

(14.15)

Exercise 14.5: Pendulum A 2 kg metal ball is suspended from a rope.

If it is released from point A and swings down to the point B (the bottom

of its arc):

1. Show that the velocity of the ball is independent of it mass.

2. Calculate the velocity of the ball at point B.

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Solution to Exercise

Step 1. • The mass of the metal ball is m = 2 kg• The change in height going from point A to point B is h =

0, 5 m• The ball is released from point A so the velocity at point, vA =

0m · s−1.

All quantities are in SI units.

Step 2. • Prove that the velocity is independent of mass.

• Find the velocity of the metal ball at point B.

Step 3. As there is no friction, mechanical energy is conserved. Therefore:

EMA= EMA

EPA+ EKA

= EPA+ EKA

mghA + 12m(vA)

2= mghB + 1

2m(vB)2

mghA + 0 = 0 + 12m(vB)

2

mghA = 12m(vB)

2

(14.16)

As the mass of the ball m appears on both sides of the equation, it

can be eliminated so that the equation becomes:

ghA =1

2(vB)

2(14.17)

2ghA = (vB)2

(14.18)

This proves that the velocity of the ball is independent of its mass.

It does not matter what its mass is, it will always have the same

velocity when it falls through this height.

Step 4. We can use the equation above, or do the calculation from ’first

principles’:

(vB)2

= 2ghA

(vB)2

= (2) (9.8) (0, 5)

(vB)2

= 9, 8

vB =√9, 8 m · s−1

(14.19)

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1.106. ENERGY GRAPHS - (NOT IN CAPS, INCLUDED FOR COMPLETENESS)CHAPTER 1. MECHANICAL ENERGY

Potential Energy

1. A tennis ball, of mass 120 g, is dropped from a height of 5 m. Ignore air friction.

a. What is the potential energy of the ball when it has fallen 3 m?

b. What is the velocity of the ball when it hits the ground?

2. A bullet, mass 50 g, is shot vertically up in the air with a muzzle velocity of 200 m · s−1. Use the Principle of

Conservation of Mechanical Energy to determine the height that the bullet will reach. Ignore air friction.

3. A skier, mass 50 kg, is at the top of a 6, 4 m ski slope.

a. Determine the maximum velocity that she can reach when she skies to the bottom of the slope.

b. Do you think that she will reach this velocity? Why/Why not?

4. A pendulum bob of mass 1, 5 kg, swings from a height A to the bottom of its arc at B. The velocity of the bob

at B is 4 m · s−1. Calculate the height A from which the bob was released. Ignore the effects of air friction.

5. Prove that the velocity of an object, in free fall, in a closed system, is independent of its mass.

Find the answers with the shortcodes:

(1.) lxV (2.) lxp (3.) lxd (4.) lxv (5.) lxw

1.106 Energy graphs - (Not in CAPS, included for completeness)

(section shortcode: P10012 )

Let us consider our example of the suitcase on the cupboard, once more.

Let’s look at each of these quantities and draw a graph for each. We will look at how each quantity changes as

the suitcase falls from the top to the bottom of the cupboard.

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CHAPTER 1. MECHANICAL ENERGY1.106. ENERGY GRAPHS - (NOT IN CAPS, INCLUDED FOR COMPLETENESS)

• Potential energy: The potential energy starts off at a maximum and decreases until it reaches zero at the

bottom of the cupboard. It had fallen a distance of 2 metres.

• Kinetic energy: The kinetic energy is zero at the start of the fall. When the suitcase reaches the ground,

the kinetic energy is a maximum. We also use distance on the x-axis.

• Mechanical energy: The mechanical energy is constant throughout the motion and is always a maximum.

At any point in time, when we add the potential energy and the kinetic energy, we will get the same number.

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1.107. SUMMARY CHAPTER 1. MECHANICAL ENERGY

1.107 Summary

(section shortcode: P10013 )

• The potential energy of an object is the energy the object has due to his position above a reference point.

• The kinetic energy of an object is the energy the object has due to its motion.

• Mechanical energy of an object is the sum of the potential energy and kinetic energy of the object.

• The unit for energy is the joule (J).

• The Law of Conservation of Energy states that energy cannot be created or destroyed, but can only be

changed from one form into another.

• The Law of Conservation of Mechanical Energy states that the total mechanical energy of an isolated system

remains constant.

• The table below summarises the most important equations:

Potential Energy EP = mgh

Kinetic Energy EK = 12mv2

Mechanical Energy EM = EK + EP

Table 14.1

1.108 End of Chapter Exercises: Mechanical Energy

(section shortcode: P10014 )

1. Give one word/term for the following descriptions.

a. The force with which the Earth attracts a body.

b. The unit for energy.

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CHAPTER 1. MECHANICAL ENERGY 1.108. END OF CHAPTER EXERCISES: MECHANICAL ENERGY

c. The movement of a body in the Earth’s gravitational field when no other forces act on it.

d. The sum of the potential and kinetic energy of a body.

e. The amount of matter an object is made up of.

2. Consider the situation where an apple falls from a tree. Indicate whether the following statements regarding

this situation are TRUE or FALSE. Write only ’true’ or ’false’. If the statement is false, write down the correct

statement.

a. The potential energy of the apple is a maximum when the apple lands on the ground.

b. The kinetic energy remains constant throughout the motion.

c. To calculate the potential energy of the apple we need the mass of the apple and the height of the tree.

d. The mechanical energy is a maximum only at the beginning of the motion.

e. The apple falls at an acceleration of 9, 8 m · s−2.

3. A man fires a rock out of a slingshot directly upward. The rock has an initial velocity of 15 m · s−1.

a. How long will it take for the rock to reach its highest point?

b. What is the maximum height that the rock will reach?

c. Draw graphs to show how the potential energy, kinetic energy and mechanical energy of the rock

changes as it moves to its highest point.

4. A metal ball of mass 200 g is tied to a light string to make a pendulum. The ball is pulled to the side to

a height (A), 10 cm above the lowest point of the swing (B). Air friction and the mass of the string can be

ignored. The ball is let go to swing freely.

a. Calculate the potential energy of the ball at point A.

b. Calculate the kinetic energy of the ball at point B.

c. What is the maximum velocity that the ball will reach during its motion?

5. A truck of mass 1, 2 tons is parked at the top of a hill, 150 m high. The truck driver lets the truck run freely

down the hill to the bottom.

a. What is the maximum velocity that the truck can achieve at the bottom of the hill?

b. Will the truck achieve this velocity? Why/why not?

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1.108. END OF CHAPTER EXERCISES: MECHANICAL ENERGY CHAPTER 1. MECHANICAL ENERGY

6. A stone is dropped from a window, 3 m above the ground. The mass of the stone is 25 g. Use the Principle

of Conservation of Energy to calculate the velocity of the stone as it reaches the ground.

Find the answers with the shortcodes:

(1.) lxf (2.) lxG (3.) lxA (4.) lxo (5.) lxs (6.) lxH

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Motion in One Dimension

1.109 Introduction

(section shortcode: P10015 )

This chapter is about how things move in a straight line or more scientifically how things move in one dimension.

This is useful for learning how to describe the movement of cars along a straight road or of trains along straight

railway tracks. If you want to understand how any object moves, for example a car on the freeway, a soccer ball

being kicked towards the goal or your dog chasing the neighbour’s cat, then you have to understand three basic

ideas about what it means when something is moving. These three ideas describe different parts of exactly how

an object moves. They are:

1. position or displacement which tells us exactly where the object is,

2. speed or velocity which tells us exactly how fast the object’s position is changing or more familiarly, how fast

the object is moving, and

3. acceleration which tells us exactly how fast the object’s velocity is changing.

You will also learn how to use position, displacement, speed, velocity and acceleration to describe the motion

of simple objects. You will learn how to read and draw graphs that summarise the motion of a moving object.

You will also learn about the equations that can be used to describe motion and how to apply these equations to

objects moving in one dimension.

1.110 Reference Point, Frame of Reference and Position

(section shortcode: P10016 )

The most important idea when studying motion, is you have to know where you are. The word position describes

your location (where you are). However, saying that you are here is meaningless, and you have to specify your

position relative to a known reference point. For example, if you are 2 m from the doorway, inside your classroom

then your reference point is the doorway. This defines your position inside the classroom. Notice that you need a

reference point (the doorway) and a direction (inside) to define your location.

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1.110. REFERENCE POINT, FRAME OF REFERENCE AND POSITIONCHAPTER 1. MOTION IN ONE DIMENSION

1.110.1 Frames of Reference

Definition: Frame of Reference

A frame of reference is a reference point combined with a set of directions.

A frame of reference is similar to the idea of a reference point. A frame of reference is defined as a reference

point combined with a set of directions. For example, a boy is standing still inside a train as it pulls out of a station.

You are standing on the platform watching the train move from left to right. To you it looks as if the boy is moving

from left to right, because relative to where you are standing (the platform), he is moving. According to the boy,

and his frame of reference (the train), he is not moving.

A frame of reference must have an origin (where you are standing on the platform) and at least a positive direction.

The train was moving from left to right, making to your right positive and to your left negative. If someone else

was looking at the same boy, his frame of reference will be different. For example, if he was standing on the other

side of the platform, the boy will be moving from right to left.

Another great example of frames of reference is a car overtaking another car on a road. Think about sitting in a

taxi passing a car. If you sit in the taxi it is your origin and the direction it is moving in will be the positive direction.

You will see the car slowly move further behind you (in the negative direction). The very important thing is that

the driver of the car has their own reference frame in which things look different. In the driver’s reference frame

the taxi is moving ahead. In one case someone is falling behind, but in the other case someone is moving ahead.

This is just a matter of perspective (from which reference frame you choose to view the situation).

For this chapter, we will only use frames of reference in the x-direction. Frames of reference will be covered in

more detail in Grade 12.

Figure 15.1: Frames of Reference

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CHAPTER 1. MOTION IN ONE DIMENSION1.110. REFERENCE POINT, FRAME OF REFERENCE AND POSITION

(Presentation: lb0 )

1.110.2 Position

Definition: Position

Position is a measurement of a location, with reference to an origin.

A position is a measurement of a location, with reference to an origin. Positions can therefore be negative or

positive. The symbol x is used to indicate position. x has units of length for example cm, m or km. Figure 15.4

shows the position of a school. Depending on what reference point we choose, we can say that the school is

300 m from Joan’s house (with Joan’s house as the reference point or origin) or 500 m from Joel’s house (with

Joel’s house as the reference point or origin).

Figure 15.4: Illustration of position

The shop is also 300m from Joan’s house, but in the opposite direction as the school. When we choose a

reference point, we have a positive direction and a negative direction. If we choose the direction towards the

school as positive, then the direction towards the shop is negative. A negative direction is always opposite to the

direction chosen as positive.

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1.110. REFERENCE POINT, FRAME OF REFERENCE AND POSITIONCHAPTER 1. MOTION IN ONE DIMENSION

Figure 15.5: The origin is at Joan’s house and the position of the school is +300 m. Positions towards the left

are defined as positive and positions towards the right are defined as negative.

Discussion : Reference Points

Divide into groups of 5 for this activity. On a straight line, choose a reference point. Since position can have both

positive and negative values, discuss the advantages and disadvantages of choosing

1. either end of the line,

2. the middle of the line.

This reference point can also be called “the origin".

Position

1. Write down the positions for objects at A, B, D and E. Do not forget the units.

2. Write down the positions for objects at F, G, H and J. Do not forget the units.

3. There are 5 houses on Newton Street, A, B, C, D and E. For all cases, assume that positions to the right

are positive.

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CHAPTER 1. MOTION IN ONE DIMENSION 1.111. DISPLACEMENT AND DISTANCE

a. Draw a frame of reference with house A as the origin and write down the positions of houses B, C, D

and E.

b. You live in house C. What is your position relative to house E?

c. What are the positions of houses A, B and D, if house B is taken as the reference point?

Find the answers with the shortcodes:

(1.) laG (2.) la7 (3.) laA

1.111 Displacement and Distance

(section shortcode: P10018 )

Definition: Displacement

Displacement is the change in an object’s position.

The displacement of an object is defined as its change in position (final position minus initial position). Displace-

ment has a magnitude and direction and is therefore a vector. For example, if the initial position of a car is xi and

it moves to a final position of xf , then the displacement is:

xf − xi (15.1)

However, subtracting an initial quantity from a final quantity happens often in Physics, so we use the shortcut ∆to mean final - initial. Therefore, displacement can be written:

∆x = xf − xi (15.2)

TIP: The symbol ∆ is read out as delta. ∆ is a letter of the Greek alphabet and is used

in Mathematics and Science to indicate a change in a certain quantity, or a final value

minus an initial value. For example, ∆x means change in x while ∆t means change

in t.

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1.111. DISPLACEMENT AND DISTANCE CHAPTER 1. MOTION IN ONE DIMENSION

TIP: The words initial and final will be used very often in Physics. Initial will always

refer to something that happened earlier in time and final will always refer to something

that happened later in time. It will often happen that the final value is smaller than the

initial value, such that the difference is negative. This is ok!

Figure 15.9: Illustration of displacement

Displacement does not depend on the path travelled, but only on the initial and final positions (Figure 15.9). We

use the word distance to describe how far an object travels along a particular path. Distance is the actual distance

that was covered. Distance (symbol d) does not have a direction, so it is a scalar. Displacement is the shortest

distance from the starting point to the endpoint – from the school to the shop in the figure. Displacement has

direction and is therefore a vector.

Figure 15.4 shows the five houses we discussed earlier. Jack walks to school, but instead of walking straight to

school, he decided to walk to his friend Joel’s house first to fetch him so that they can walk to school together.

Jack covers a distance of 400 m to Joel’s house and another 500 m to school. He covers a distance of 900 m. His

displacement, however, is only 100 m towards the school. This is because displacement only looks at the starting

position (his house) and the end position (the school). It does not depend on the path he travelled.

To calculate his distance and displacement, we need to choose a reference point and a direction. Let’s choose

Jack’s house as the reference point, and towards Joel’s house as the positive direction (which means that towards

the school is negative). We would do the calculations as follows:

Distance (D) = path travelled

= 400 m + 500 m

= 900 m

(15.3)

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CHAPTER 1. MOTION IN ONE DIMENSION 1.111. DISPLACEMENT AND DISTANCE

Displacement (∆x) = xf − xi

= −100 m + 0 m

= −100 m

(15.4)

NOTE: You may also see d used for distance. We will use D in this book, but you may

see d used in other books.

Joel walks to school with Jack and after school walks back home. What is Joel’s displacement and what distance

did he cover? For this calculation we use Joel’s house as the reference point. Let’s take towards the school as

the positive direction.

Distance (D) = path travelled

= 500 m + 500 m

= 1000 m

(15.5)

Displacement (∆x) = xf − xi

= 0 m+ 0 m

= 0 m

(15.6)

It is possible to have a displacement of 0 m and a distance that is not 0 m. This happens when an object

completes a round trip back to its original position, like an athlete running around a track.

1.111.1 Interpreting Direction

Very often in calculations you will get a negative answer. For example, Jack’s displacement in the example

above, is calculated as −100 m. The minus sign in front of the answer means that his displacement is 100 min the opposite direction (opposite to the direction chosen as positive in the beginning of the question). When

we start a calculation we choose a frame of reference and a positive direction. In the first example above, the

reference point is Jack’s house and the positive direction is towards Joel’s house. Therefore Jack’s displacement

is 100 m towards the school. Notice that distance has no direction, but displacement has.

1.111.2 Differences between Distance and Displacement

Definition: Vectors and Scalars

A vector is a physical quantity with magnitude (size) and direction. A scalar is a physical

quantity with magnitude (size) only.

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1.112. SPEED, AVERAGE VELOCITY AND INSTANTANEOUS VELOCITYCHAPTER 1. MOTION IN ONE DIMENSION

The differences between distance and displacement can be summarised as:

Distance Displacement

1. depends on the path 1. independent of path taken

2. always positive 2. can be positive or negative

3. is a scalar 3. is a vector

4. does not have a direction 4. has a direction

Table 15.1

Point of Reference

1. Use Figure 15.4 to answer the following questions.

a. Jill walks to Joan’s house and then to school, what is her distance and displacement?

b. John walks to Joan’s house and then to school, what is his distance and displacement?

c. Jack walks to the shop and then to school, what is his distance and displacement?

d. What reference point did you use for each of the above questions?

2. You stand at the front door of your house (displacement, ∆x = 0m). The street is 10 m away from the front

door. You walk to the street and back again.

a. What is the distance you have walked?

b. What is your final displacement?

c. Is displacement a vector or a scalar? Give a reason for your answer.

Find the answers with the shortcodes:

(1.) lao (2.) las

1.112 Speed, Average Velocity and Instantaneous Velocity

(section shortcode: P10019 )

Definition: Velocity

Velocity is the rate of change of displacement.

Definition: Instantaneous velocity

Instantaneous velocity is the velocity of a body at a specific instant in time.

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CHAPTER 1. MOTION IN ONE DIMENSION1.112. SPEED, AVERAGE VELOCITY AND INSTANTANEOUS VELOCITY

Definition: Average velocity

Average velocity is the total displacement of a body over a time interval.

Velocity is the rate of change of position. It tells us how much an object’s position changes in time. This is the

same as the displacement divided by the time taken. Since displacement is a vector and time taken is a scalar,

velocity is also a vector. We use the symbol v for velocity. If we have a displacement of ∆x and a time taken of

∆t, v is then defined as:

velocity(

in m · s−1)

= change in displacement (in m)change in time (in s)

v = ∆x∆t

(15.7)

Velocity can be positive or negative. Positive values of velocity mean that the object is moving away from the

reference point or origin and negative values mean that the object is moving towards the reference point or origin.

TIP: An instant in time is different from the time taken or the time interval. It is therefore

useful to use the symbol t for an instant in time (for example during the 4th second) and

the symbol ∆t for the time taken (for example during the first 5 seconds of the motion).

Average velocity (symbol v) is the displacement for the whole motion divided by the time taken for the whole

motion. Instantaneous velocity is the velocity at a specific instant in time.

This is terminology that occurs quite often and it is important to always remember that instantaneous and average

quantities are not always the same. In fact, they can be very different. The magnitude of the instantaneous veloc-

ity is the same as the slope of the line which is a tangent to the displacement curve at the time of measurement.

The magnitude of the average velocity is the same as the slope of the line between the start and end points of

the interval.

If you want to think about why the tangent at a particular time gives us the velocity remember that the velocity

is the slope of the displacement curve. Now, a simple why to start thinking about it is to image you could zoom

(magnify) the displacement curve at the point. The more you zoom in the more it will look like a straight line at

the point and that straight line will be very similar to the tangent line at the point. This isn’t a mathematical proof

but you will learn one later on in mathematics about limits and slopes.

(Average) Speed (symbol s) is the distance travelled (D) divided by the time taken (∆t) for the journey. Distance

and time are scalars and therefore speed will also be a scalar. Speed is calculated as follows:

speed(

in m · s−1)

=distance (in m)

time (in s)(15.8)

s =D

∆t(15.9)

Instantaneous speed is the magnitude of instantaneous velocity. It has the same value, but no direction.

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1.112. SPEED, AVERAGE VELOCITY AND INSTANTANEOUS VELOCITYCHAPTER 1. MOTION IN ONE DIMENSION

Exercise 15.1: Average speed and average velocity James walks

2 km away from home in 30 minutes. He then turns around and walks

back home along the same path, also in 30 minutes. Calculate James’

average speed and average velocity.

Solution to Exercise

Step 1. The question explicitly gives

• the distance and time out (2 km in 30 minutes)

• the distance and time back (2 km in 30 minutes)

Step 2. The information is not in SI units and must therefore be converted.

To convert km to m, we know that:

1 km = 1 000 m

∴ 2 km = 2 000 m (multiply both sides by2)(15.10)

Similarly, to convert 30 minutes to seconds,

1 min = 60 s

∴ 30 min = 1 800 s (multiply both sides by30)(15.11)

Step 3. James started at home and returned home, so his displacement is

0 m.

∆x = 0 m (15.12)

James walked a total distance of 4 000 m (2 000 m out and 2 000 m

back).

D = 4 000 m (15.13)

Step 4. James took 1 800 s to walk out and 1 800 s to walk back.

∆t = 3 600 s (15.14)

Step 5.

s = D∆t

= 4 000 m3 600 s

= 1, 11 m · s−1

(15.15)

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CHAPTER 1. MOTION IN ONE DIMENSION1.112. SPEED, AVERAGE VELOCITY AND INSTANTANEOUS VELOCITY

Step 6.

v = ∆x∆t

= 0 m3 600 s

= 0 m · s−1

(15.16)

Exercise 15.2: Instantaneous Speed and Velocity A man runs

around a circular track of radius 100 m. It takes him 120 s to complete

a revolution of the track. If he runs at constant speed, calculate:

1. his speed,

2. his instantaneous velocity at point A,

3. his instantaneous velocity at point B,

4. his average velocity between points A and B,

5. his average speed during a revolution.

6. his average velocity during a revolution.

Solution to Exercise

Step 1. To determine the man’s speed we need to know the distance he

travels and how long it takes. We know it takes 120 s to complete

one revolution of the track.(A revolution is to go around the track

once.)

Step 2. What distance is one revolution of the track? We know the track is

a circle and we know its radius, so we can determine the distance

around the circle. We start with the equation for the circumference

of a circle

C = 2πr

= 2π (100m)

= 628, 32m

(15.17)

Therefore, the distance the man covers in one revolution is

628, 32 m.

Step 3. We know that speed is distance covered per unit time. So if we

divide the distance covered by the time it took we will know how

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much distance was covered for every unit of time. No direction is

used here because speed is a scalar.

s = D∆t

= 628,32 m120 s

= 5, 24m · s−1

(15.18)

Step 4. Consider the point A in the diagram. We know which way the man

is running around the track and we know his speed. His velocity

at point A will be his speed (the magnitude of the velocity) plus his

direction of motion (the direction of his velocity). The instant that he

arrives at A he is moving as indicated in the diagram. His velocity

will be 5, 24 m · s−1 West.

Step 5. Consider the point B in the diagram. We know which way the man

is running around the track and we know his speed. His velocity

at point B will be his speed (the magnitude of the velocity) plus his

direction of motion (the direction of his velocity). The instant that he

arrives at B he is moving as indicated in the diagram. His velocity

will be 5, 24 m · s−1 South.

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Step 6. To determine the average velocity between A and B, we need the

change in displacement between A and B and the change in time

between A and B. The displacement from A and B can be calculated

by using the Theorem of Pythagoras:

(∆x)2

= (100m)2+ (100m)

2

= 20000m

∆x = 141, 42135... m

(15.19)

The time for a full revolution is 120 s, therefore the time for a 14 of

a revolution is 30 s.

vAB = ∆x∆t

= 141,42...m30s

= 4.71 m · s−1

(15.20)

Velocity is a vector and needs a direction.

Triangle AOB is isosceles and therefore angle BAO = 45.

The direction is between west and south and is therefore southwest.

The final answer is: v = 4.71 m · s−1, southwest.

Step 7. Because he runs at a constant rate, we know that his speed any-

where around the track will be the same. His average speed is

5, 24 m · s−1 .

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Step 8.

TIP: Remember - displacement can be zero even when dis-

tance travelled is not!

To calculate average velocity we need his total displacement and

his total time. His displacement is zero because he ends up where

he started. His time is 120 s. Using these we can calculate his

average velocity:

v = ∆x∆t

= 0 m120 s

= 0 m · s−1

(15.21)

1.112.1 Differences between Speed and Velocity

The differences between speed and velocity can be summarised as:

Speed Velocity

1. depends on the path taken 1. independent of path taken

2. always positive 2. can be positive or negative

3. is a scalar 3. is a vector

4. no dependence on direction and so is only positive 4. direction can be guessed from the sign (i.e. posi-

tive or negative)

Table 15.2

Additionally, an object that makes a round trip, i.e. travels away from its starting point and then returns to the

same point has zero velocity but travels a non-zero speed.

Displacement and related quantities

1. Theresa has to walk to the shop to buy some milk. After walking 100 m, she realises that she does not

have enough money, and goes back home. If it took her two minutes to leave and come back, calculate the

following:

a. How long was she out of the house (the time interval ∆t in seconds)?

b. How far did she walk (distance (D))?

c. What was her displacement (∆x)?

d. What was her average velocity (in m·s−1)?

e. What was her average speed (in m·s−1)?

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2. Desmond is watching a straight stretch of road from his classroom window. He can see two poles which he

earlier measured to be 50 m apart. Using his stopwatch, Desmond notices that it takes 3 s for most cars to

travel from the one pole to the other.

a. Using the equation for velocity (v = ∆x∆t

), show all the working needed to calculate the velocity of a car

travelling from the left to the right.

b. If Desmond measures the velocity of a red Golf to be −16, 67 m · s−1, in which direction was the Gold

travelling? Desmond leaves his stopwatch running, and notices that at t = 5, 0 s, a taxi passes the left

pole at the same time as a bus passes the right pole. At time t = 7, 5 s the taxi passes the right pole.

At time t = 9, 0 s, the bus passes the left pole.

c. How long did it take the taxi and the bus to travel the distance between the poles? (Calculate the time

interval (∆t) for both the taxi and the bus).

d. What was the velocity of the taxi and the bus?

e. What was the speed of the taxi and the bus?

f. What was the speed of taxi and the bus in km · h−1?

3. A rabbit runs across a freeway. There is a car, 100 m away travelling towards the rabbit.

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a. If the car is travelling at 120 km · h−1, what is the car’s speed in m · s−1.

b. How long will it take the a car to travel 100 m?

c. If the rabbit is running at 10 km · h−1, what is its speed in m · s−1?

d. If the freeway has 3 lanes, and each lane is 3 m wide, how long will it take for the rabbit to cross all

three lanes?

e. If the car is travelling in the furthermost lane from the rabbit, will the rabbit be able to cross all 3 lanes

of the freeway safely?

Find the answers with the shortcodes:

(1.) laH (2.) la6 (3.) laF

Investigation : An Exercise in Safety

Divide into groups of 4 and perform the following investigation. Each group will be performing the same investi-

gation, but the aim for each group will be different.

1. Choose an aim for your investigation from the following list and formulate a hypothesis:

• Do cars travel at the correct speed limit?

• Is is safe to cross the road outside of a pedestrian crossing?

• Does the colour of your car determine the speed you are travelling at?

• Any other relevant question that you would like to investigate.

2. On a road that you often cross, measure out 50 m along a straight section, far away from traffic lights or

intersections.

3. Use a stopwatch to record the time each of 20 cars take to travel the 50 m section you measured.

4. Design a table to represent your results. Use the results to answer the question posed in the aim of the

investigation. You might need to do some more measurements for your investigation. Plan in your group

what else needs to be done.

5. Complete any additional measurements and write up your investigation under the following headings:

• Aim and Hypothesis

• Apparatus

• Method

• Results

• Discussion

• Conclusion

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6. Answer the following questions:

a. How many cars took less than 3 s to travel 50 m?

b. What was the shortest time a car took to travel 50 m?

c. What was the average time taken by the 20 cars?

d. What was the average speed of the 20 cars?

e. Convert the average speed to km · h−1.

1.113 Acceleration

(section shortcode: P10020 )

Definition: Acceleration

Acceleration is the rate of change of velocity.

Acceleration (symbol a) is the rate of change of velocity. It is a measure of how fast the velocity of an object

changes in time. If we have a change in velocity (∆v) over a time interval (∆t), then the acceleration (a) is

defined as:

acceleration(

in m · s−2)

=change in velocity

(

in m · s−1)

change in time (in s)(15.22)

a =∆v

∆t(15.23)

Since velocity is a vector, acceleration is also a vector. Acceleration does not provide any information about a

motion, but only about how the motion changes. It is not possible to tell how fast an object is moving or in which

direction from the acceleration.

Like velocity, acceleration can be negative or positive. We see that when the sign of the acceleration and the

velocity are the same, the object is speeding up. If both velocity and acceleration are positive, the object is

speeding up in a positive direction. If both velocity and acceleration are negative, the object is speeding up in a

negative direction. If velocity is positive and acceleration is negative, then the object is slowing down. Similarly,

if the velocity is negative and the acceleration is positive the object is slowing down. This is illustrated in the

following worked example.

Exercise 15.3: Acceleration A car accelerates uniformly from an initial

velocity of 2 m·s−1 to a final velocity of 10 m·s1 in 8 seconds. It then

slows down uniformly to a final velocity of 4 m·s−1 in 6 seconds. Calcu-

late the acceleration of the car during the first 8 seconds and during the

last 6 seconds.

Solution to Exercise

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Step 1. Consider the motion of the car in two parts: the first 8 seconds and

the last 6 seconds.

For the first 8 seconds:

vi = 2 m · s−1

vf = 10 m · s−1

ti = 0 s

tf = 8 s

(15.24)

For the last 6 seconds:

vi = 10 m · s−1

vf = 4 m · s−1

ti = 8 s

tf = 14 s

(15.25)

Step 2. For the first 8 seconds:

a = ∆v∆t

= 10m·s−1−2m·s−1

8s−0s

= 1 m · s−2

(15.26)

For the next 6 seconds:

a = ∆v∆t

= 4m·s−1−10m·s−1

14s−8s

= −1 m · s−2

(15.27)

During the first 8 seconds the car had a positive acceleration. This

means that its velocity increased. The velocity is positive so the car

is speeding up. During the next 6 seconds the car had a negative

acceleration. This means that its velocity decreased. The velocity

is negative so the car is slowing down.

TIP: Acceleration does not tell us about the direction of the motion. Acceleration only

tells us how the velocity changes.

TIP: Avoid the use of the word deceleration to refer to a negative acceleration. This

word usually means slowing down and it is possible for an object to slow down with

both a positive and negative acceleration, because the sign of the velocity of the object

must also be taken into account to determine whether the body is slowing down or not.

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1.113.1 Acceleration

1. An athlete is accelerating uniformly from an initial velocity of 0 m·s−1to a final velocity of 4 m·s−1in 2

seconds. Calculate his acceleration. Let the direction that the athlete is running in be the positive direction.

2. A bus accelerates uniformly from an initial velocity of 15 m·s−1to a final velocity of 7 m·s−1in 4 seconds.

Calculate the acceleration of the bus. Let the direction of motion of the bus be the positive direction.

3. An aeroplane accelerates uniformly from an initial velocity of 200 m·s−1to a velocity of 100 m·s−1in 10

seconds. It then accelerates uniformly to a final velocity of 240 m·s−1in 20 seconds. Let the direction of

motion of the aeroplane be the positive direction.

a. Calculate the acceleration of the aeroplane during the first 10 seconds of the motion.

b. Calculate the acceleration of the aeroplane during the next 14 seconds of its motion.

The following video provides a summary of distance, velocity and acceleration. Note that in this video a different

convention for writing units is used. You should not use this convention when writing units in physics.

Khan academy video on motion - 1 (Video: P10021)

Find the answers with the shortcodes:

(1.) l1k (2.) l10 (3.) l18

1.114 Description of Motion

(section shortcode: P10022 )

The purpose of this chapter is to describe motion, and now that we understand the definitions of displacement,

distance, velocity, speed and acceleration, we are ready to start using these ideas to describe how an object is

moving. There are many ways of describing motion:

1. words

2. diagrams

3. graphs

These methods will be described in this section.

We will consider three types of motion: when the object is not moving (stationary object), when the object is

moving at a constant velocity (uniform motion) and when the object is moving at a constant acceleration (motion

at constant acceleration).

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1.114.1 Stationary Object

The simplest motion that we can come across is that of a stationary object. A stationary object does not move and

so its position does not change, for as long as it is standing still. An example of this situation is when someone is

waiting for something without moving. The person remains in the same position.

Lesedi is waiting for a taxi. He is standing two metres from a stop street at t = 0 s. After one minute, at t = 60

s, he is still 2 metres from the stop street and after two minutes, at t = 120 s, also 2 metres from the stop street.

His position has not changed. His displacement is zero (because his position is the same), his velocity is zero

(because his displacement is zero) and his acceleration is also zero (because his velocity is not changing).

We can now draw graphs of position vs. time (x vs. t), velocity vs. time (v vs. t) and acceleration vs. time (avs. t) for a stationary object. The graphs are shown in Figure 15.22. Lesedi’s position is 2 metres from the stop

street. If the stop street is taken as the reference point, his position remains at 2 metres for 120 seconds. The

graph is a horizontal line at 2 m. The velocity and acceleration graphs are also shown. They are both horizontal

lines on the x-axis. Since his position is not changing, his velocity is 0 m · s−1 and since velocity is not changing,

acceleration is 0 m · s−2.

Figure 15.22: Graphs for a stationary object (a) position vs. time (b) velocity vs. time (c) acceleration vs. time.

Definition: Gradient

The gradient of a line can be calculated by dividing the change in the y-value by the change

in the x-value.

m = ∆y∆x

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Since we know that velocity is the rate of change of position, we can confirm the value for the velocity vs. time

graph, by calculating the gradient of the x vs. t graph.

TIP: The gradient of a position vs. time graph gives the velocity.

If we calculate the gradient of the x vs. t graph for a stationary object we get:

v = ∆x∆t

=xf−xi

tf−ti

= 2 m−2 m120 s−60 s (initial position = final position)

= 0 m · s−1 (for the time that Lesedi is stationary)

(15.28)

Similarly, we can confirm the value of the acceleration by calculating the gradient of the velocity vs. time graph.

TIP: The gradient of a velocity vs. time graph gives the acceleration.

If we calculate the gradient of the v vs. t graph for a stationary object we get:

a = ∆v∆t

=vf−vi

tf−ti

= 0 m·s−1−0 m·s−1

120 s−60 s

= 0 m · s−2

(15.29)

Additionally, because the velocity vs. time graph is related to the position vs. time graph, we can use the area

under the velocity vs. time graph to calculate the displacement of an object.

TIP: The area under the velocity vs. time graph gives the displacement.

The displacement of the object is given by the area under the graph, which is 0 m. This is obvious, because the

object is not moving.

1.114.2 Motion at Constant Velocity

Motion at a constant velocity or uniform motion means that the position of the object is changing at the same rate.

Assume that Lesedi takes 100 s to walk the 100 m to the taxi-stop every morning. If we assume that Lesedi’s

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house is the origin, then Lesedi’s velocity is:

v = ∆x∆t

=xf−xi

tf−ti

= 100 m−0 m100 s−0 s

= 1 m · s−1

(15.30)

Lesedi’s velocity is 1 m·s−1. This means that he walked 1 m in the first second, another metre in the second

second, and another in the third second, and so on. For example, after 50 s he will be 50 m from home. His

position increases by 1 m every 1 s. A diagram of Lesedi’s position is shown in Figure 15.23.

Figure 15.23: Diagram showing Lesedi’s motion at a constant velocity of 1 m·s−1

We can now draw graphs of position vs.time (x vs. t), velocity vs.time (v vs. t) and acceleration vs.time (a vs. t)for Lesedi moving at a constant velocity. The graphs are shown in Figure 15.24.

Figure 15.24: Graphs for motion at constant velocity (a) position vs. time (b) velocity vs. time (c) acceleration

vs. time. The area of the shaded portion in the v vs. t graph corresponds to the object’s displacement.

In the evening Lesedi walks 100 m from the bus stop to his house in 100 s. Assume that Lesedi’s house is the

origin. The following graphs can be drawn to describe the motion.

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Figure 15.25: Graphs for motion with a constant negative velocity (a) position vs. time (b) velocity vs. time (c)

acceleration vs. time. The area of the shaded portion in the v vs.t graph corresponds to the object’s displace-

ment.

We see that the v vs. t graph is a horisontal line. If the velocity vs. time graph is a horisontal line, it means that

the velocity is constant (not changing). Motion at a constant velocity is known as uniform motion.

We can use the x vs. t to calculate the velocity by finding the gradient of the line.

v = ∆x∆t

=xf−xi

tf−ti

= 0 m−100 m100 s−0 s

= −1 m · s−1

(15.31)

Lesedi has a velocity of −1 m · s−1, or 1 m · s−1 towards his house. You will notice that the v vs. t graph is a

horisontal line corresponding to a velocity of −1 m · s−1. The horizontal line means that the velocity stays the

same (remains constant) during the motion. This is uniform velocity.

We can use the v vs. t to calculate the acceleration by finding the gradient of the line.

a = ∆v∆t

=vf−vi

tf−ti

= 1 m·s−1−1 m·s−1

100 s−0 s

= 0 m · s−2

(15.32)

Lesedi has an acceleration of 0 m · s−2. You will notice that the graph of a vs.t is a horisontal line corresponding

to an acceleration value of 0 m · s−2. There is no acceleration during the motion because his velocity does not

change.

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We can use the v vs. t to calculate the displacement by finding the area under the graph.

v = Area under graph

= ℓ× b

= 100× (−1)

= −100 m

(15.33)

This means that Lesedi has a displacement of 100 m towards his house.

Velocity and acceleration

1. Use the graphs in Figure 15.24 to calculate each of the following:

a. Calculate Lesedi’s velocity between 50 s and 100 s using the x vs. t graph. Hint: Find the gradient of

the line.

b. Calculate Lesedi’s acceleration during the whole motion using the v vs. t graph.

c. Calculate Lesedi’s displacement during the whole motion using the v vs. t graph.

2. Thandi takes 200 s to walk 100 m to the bus stop every morning. In the evening Thandi takes 200 s to walk

100 m from the bus stop to her home.

a. Draw a graph of Thandi’s position as a function of time for the morning (assuming that Thandi’s home

is the reference point). Use the gradient of the x vs. t graph to draw the graph of velocity vs. time. Use

the gradient of the v vs. t graph to draw the graph of acceleration vs. time.

b. Draw a graph of Thandi’s position as a function of time for the evening (assuming that Thandi’s home is

the origin). Use the gradient of the x vs. t graph to draw the graph of velocity vs. time. Use the gradient

of the v vs. t graph to draw the graph of acceleration vs. time.

c. Discuss the differences between the two sets of graphs in questions 2 and 3.

Find the answers with the shortcodes:

(1.) l19 (2.) l19

Experiment : Motion at constant velocity

Aim: To measure the position and time during motion at constant velocity and determine the average velocity as

the gradient of a “Position vs. Time" graph.

Apparatus: A battery operated toy car, stopwatch, meter stick or measuring tape.

Method

1. Work with a friend. Copy the table below into your workbook.

2. Complete the table by timing the car as it travels each distance.

3. Time the car twice for each distance and take the average value as your accepted time.

4. Use the distance and average time values to plot a graph of “Distance vs. Time" onto graph paper. Stick

the graph paper into your workbook. (Remember that “A vs. B" always means “y vs. x").

5. Insert all axis labels and units onto your graph.

6. Draw the best straight line through your data points.

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7. Find the gradient of the straight line. This is the average velocity.

Results:

Distance (m) Time (s)

1 2 Ave.

0

0,5

1,0

1,5

2,0

2,5

3,0

Table 15.3

Conclusions: Answer the following questions in your workbook:

1. Did the car travel with a constant velocity?

2. How can you tell by looking at the “Distance vs. Time" graph if the velocity is constant?

3. How would the “Distance vs. Time" look for a car with a faster velocity?

4. How would the “Distance vs. Time" look for a car with a slower velocity?

1.114.3 Motion at Constant Acceleration

The final situation we will be studying is motion at constant acceleration. We know that acceleration is the rate

of change of velocity. So, if we have a constant acceleration, this means that the velocity changes at a constant

rate.

Let’s look at our first example of Lesedi waiting at the taxi stop again. A taxi arrived and Lesedi got in. The taxi

stopped at the stop street and then accelerated as follows: After 1 s the taxi covered a distance of 2, 5 m, after

2 s it covered 10 m, after 3 s it covered 22, 5 m and after 4 s it covered 40 m. The taxi is covering a larger distance

every second. This means that it is accelerating.

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To calculate the velocity of the taxi you need to calculate the gradient of the line at each second:

v1s = ∆x∆t

=xf−xi

tf−ti

= 5 m−0 m1,5 s−0,5 s

= 5 m · s−1

(15.34)

v2s = ∆x∆t

=xf−xi

tf−ti

= 15 m−5 m2,5 s−1,5 s

= 10 m · s−1

(15.35)

v3s = ∆x∆t

=xf−xi

tf−ti

= 30 m−15 m3,5 s−2,5 s

= 15 m · s−1

(15.36)

From these velocities, we can draw the velocity-time graph which forms a straight line.

The acceleration is the gradient of the v vs. t graph and can be calculated as follows:

a = ∆v∆t

=vf−vi

tf−ti

= 15 m·s−1−5 m·s−1

3 s−1 s

= 5 m · s−2

(15.37)

The acceleration does not change during the motion (the gradient stays constant). This is motion at constant or

uniform acceleration.

The graphs for this situation are shown in Figure 15.27.

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Figure 15.27: Graphs for motion with a constant acceleration (a) position vs. time (b) velocity vs. time (c)

acceleration vs. time.

Velocity from Acceleration vs. Time Graphs

Just as we used velocity vs. time graphs to find displacement, we can use acceleration vs. time graphs to find

the velocity of an object at a given moment in time. We simply calculate the area under the acceleration vs. time

graph, at a given time. In the graph below, showing an object at a constant positive acceleration, the increase in

velocity of the object after 2 seconds corresponds to the shaded portion.

v = area of rectangle = a×∆t

= 5 m · s−2 × 2 s

= 10m · s−1

(15.38)

The velocity of the object at t = 2s is therefore 10 m · s−1. This corresponds with the values obtained in

Figure 15.27.

1.114.4 Summary of Graphs

The relation between graphs of position, velocity and acceleration as functions of time is summarised in Fig-

ure 15.28.

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Figure 15.28: Position-time, velocity-time and acceleration-time graphs.

TIP: Often you will be required to describe the motion of an object that is presented as

a graph of either position, velocity or acceleration as functions of time. The description

of the motion represented by a graph should include the following (where possible):

1.whether the object is moving in the positive or negative direction

2.whether the object is at rest, moving at constant velocity or moving at constant

positive acceleration (speeding up) or constant negative acceleration (slowing

down)

You will also often be required to draw graphs based on a description of the motion in

words or from a diagram. Remember that these are just different methods of presenting

the same information. If you keep in mind the general shapes of the graphs for the

different types of motion, there should not be any difficulty with explaining what is

happening.

1.114.5 Experiment: Position versus time using a ticker timer

Aim: To measure the position and time during motion and to use that data to plot a “Position vs. Time" graph.

Apparatus: Trolley, ticker tape apparatus, tape, graph paper, ruler, ramp

Method:

1. Work with a friend. Copy the table below into your workbook.

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2. Attach a length of tape to the trolley.

3. Run the other end of the tape through the ticker timer.

4. Start the ticker timer going and roll the trolley down the ramp.

5. Repeat steps 1 - 3.

6. On each piece of tape, measure the distance between successive dots. Note these distances in the table

below.

7. Use the frequency of the ticker timer to work out the time intervals between successive dots. Note these

times in the table below,

8. Work out the average values for distance and time.

9. Use the average distance and average time values to plot a graph of “Distance vs. Time" onto graph paper.

Stick the graph paper into your workbook. (Remember that “A vs. B" always means “y vs. x").

10. Insert all axis labels and units onto your graph.

11. Draw the best straight line through your data points.

Results:

Distance (m) Time (s)

1 2 Ave. 1 2 Ave.

Table 15.4

Discussion: Describe the motion of the trolley down the ramp.

1.114.6 Worked Examples

The worked examples in this section demonstrate the types of questions that can be asked about graphs.

Exercise 15.4: Description of motion based on a position-time

graph The position vs. time graph for the motion of a car is given be-

low. Draw the corresponding velocity vs. time and acceleration vs. time

graphs, and then describe the motion of the car.

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Solution to Exercise

Step 1. The question gives a position vs. time graph and the following three

things are required:

a. Draw a v vs. t graph.

b. Draw an a vs. t graph.

c. Describe the motion of the car.

To answer these questions, break the motion up into three sections:

0 – 2 seconds, 2 – 4 seconds and 4 – 6 seconds.

Step 2. For the first 2 seconds we can see that the displacement remains

constant - so the object is not moving, thus it has zero velocity dur-

ing this time. We can reach this conclusion by another path too:

remember that the gradient of a displacement vs. time graph is the

velocity. For the first 2 seconds we can see that the displacement

vs. time graph is a horizontal line, ie. it has a gradient of zero. Thus

the velocity during this time is zero and the object is stationary.

Step 3. For the next 2 seconds, displacement is increasing with time so

the object is moving. Looking at the gradient of the displacement

graph we can see that it is not constant. In fact, the slope is get-

ting steeper (the gradient is increasing) as time goes on. Thus,

remembering that the gradient of a displacement vs. time graph is

the velocity, the velocity must be increasing with time during this

phase.

Step 4. For the final 2 seconds we see that displacement is still increasing

with time, but this time the gradient is constant, so we know that the

object is now travelling at a constant velocity, thus the velocity vs.

time graph will be a horizontal line during this stage. We can now

draw the graphs:

So our velocity vs. time graph looks like this one below. Because

we haven’t been given any values on the vertical axis of the dis-

placement vs. time graph, we cannot figure out what the exact

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gradients are and therefore what the values of the velocities are. In

this type of question it is just important to show whether velocities

are positive or negative, increasing, decreasing or constant.

Once we have the velocity vs. time graph its much easier to get

the acceleration vs. time graph as we know that the gradient of a

velocity vs. time graph is the just the acceleration.

Step 5. For the first 2 seconds the velocity vs. time graph is horisontal and

has a value of zero, thus it has a gradient of zero and there is no

acceleration during this time. (This makes sense because we know

from the displacement time graph that the object is stationary during

this time, so it can’t be accelerating).

Step 6. For the next 2 seconds the velocity vs. time graph has a positive

gradient. This gradient is not changing (i.e. its constant) throughout

these 2 seconds so there must be a constant positive acceleration.

Step 7. For the final 2 seconds the object is traveling with a constant ve-

locity. During this time the gradient of the velocity vs. time graph

is once again zero, and thus the object is not accelerating. The

acceleration vs. time graph looks like this:

Step 8. A brief description of the motion of the object could read something

like this: At t = 0 s and object is stationary at some position and

remains stationary until t = 2 s when it begins accelerating. It

accelerates in a positive direction for 2 seconds until t = 4 s and

then travels at a constant velocity for a further 2 seconds.

Exercise 15.5: Calculations from a velocity vs. time graph The

velocity vs. time graph of a truck is plotted below. Calculate the distance

and displacement of the truck after 15 seconds.

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Solution to Exercise

Step 1. We are asked to calculate the distance and displacement of the

car. All we need to remember here is that we can use the area

between the velocity vs. time graph and the time axis to determine

the distance and displacement.

Step 2. Break the motion up: 0 – 5 seconds, 5 – 12 seconds, 12 – 14

seconds and 14 – 15 seconds.

For 0 – 5 seconds: The displacement is equal to the area of the

triangle on the left:

Area = 12 b× h

= 12 × 5 s × 4m · s−1

= 10 m

(15.39)

For 5 – 12 seconds: The displacement is equal to the area of the

rectangle:

Area = ℓ× b

= 7 s × 4m · s−1

= 28 m2

(15.40)

For 12 – 14 seconds the displacement is equal to the area of the

triangle above the time axis on the right:

Area = 12 b× h

= 12 × 2 s × 4m · s−1

= 4 m

(15.41)

For 14 – 15 seconds the displacement is equal to the area of the

triangle below the time axis:

Area = 12 b× h

= 12 × 1 s × 2m · s−1

= 1 m

(15.42)

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Step 3. Now the total distance of the car is the sum of all of these areas:

∆x = 10 m + 28 m + 4 m+ 1 m

= 43 m(15.43)

Step 4. Now the total displacement of the car is just the sum of all of these

areas. HOWEVER, because in the last second (from t = 14 s to

t = 15 s) the velocity of the car is negative, it means that the car was

going in the opposite direction, i.e. back where it came from! So, to

find the total displacement, we have to add the first 3 areas (those

with positive displacements) and subtract the last one (because it

is a displacement in the opposite direction).

∆x = 10 m + 28 m + 4 m− 1 m

= 41min the positive direction(15.44)

Exercise 15.6: Velocity from a position vs. time graph The position

vs. time graph below describes the motion of an athlete.

1. What is the velocity of the athlete during the first 4 seconds?

2. What is the velocity of the athlete from t = 4 s to t = 7 s?

Solution to Exercise

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Step 1. The velocity is given by the gradient of a position vs. time graph.

During the first 4 seconds, this is

v = ∆x∆t

= 4 m−0 m4 s−0 s

= 1 m · s−1

(15.45)

Step 2. For the last 3 seconds we can see that the displacement stays con-

stant. The graph shows a horisontal line and therefore the gradient

is zero. Thus v = 0 m · s−1.

Exercise 15.7: Drawing a v vs. t graph from an a vs. t graph The

acceleration vs. time graph for a car starting from rest, is given below.

Calculate the velocity of the car and hence draw the velocity vs. time

graph.

Solution to Exercise

Step 1. The motion of the car can be divided into three time sections: 0 – 2

seconds; 2 – 4 seconds and 4 – 6 seconds. To be able to draw the

velocity vs. time graph, the velocity for each time section needs to

be calculated. The velocity is equal to the area of the square under

the graph:

For 0 – 2 seconds:

Area = ℓ× b

= 2 s × 2m · s−2

= 4 m · s−1

(15.46)

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The velocity of the car is 4 m·s−1 at t = 2s. For 2 – 4 seconds:

Area = ℓ× b

= 2 s × 0m · s−2

= 0 m · s−1

(15.47)

The velocity of the car is 0 m·s−1 from t = 2 s to t = 4 s. For 4 – 6

seconds:

Area = ℓ× b

= 2 s ×−2m · s−2

= −4 m · s−1

(15.48)

The acceleration had a negative value, which means that the ve-

locity is decreasing. It starts at a velocity of 4 m·s−1 and decreases

to 0 m·s−1.

Step 2. The velocity vs. time graph looks like this:

1.114.7 Graphs

1. A car is parked 10 m from home for 10 minutes. Draw a displacement-time, velocity-time and acceleration-

time graphs for the motion. Label all the axes.

2. A bus travels at a constant velocity of 12 m · s−1for 6 seconds. Draw the displacement-time, velocity-time

and acceleration-time graph for the motion. Label all the axes.

3. An athlete runs with a constant acceleration of 1 m · s−2 for 4 s. Draw the acceleration-time, velocity-time

and displacement time graphs for the motion. Accurate values are only needed for the acceleration-time

and velocity-time graphs.

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1.115. EQUATIONS OF MOTION CHAPTER 1. MOTION IN ONE DIMENSION

4. The following velocity-time graph describes the motion of a car. Draw the displacement-time graph and the

acceleration-time graph and explain the motion of the car according to the three graphs.

5. The following velocity-time graph describes the motion of a truck. Draw the displacement-time graph and

the acceleration-time graph and explain the motion of the truck according to the three graphs.

This simulation allows you the opportunity to plot graphs of motion and to see how the graphs of motion change

when you move the man. (Simulation: lb8)

Find the answers with the shortcodes:

(1.) l1I (2.) l15 (3.) l1N (4.) l1R (5.) l1n

1.115 Equations of Motion

(section shortcode: P10023 )

In this chapter we will look at the third way to describe motion. We have looked at describing motion in terms of

graphs and words. In this section we examine equations that can be used to describe motion.

This section is about solving problems relating to uniformly accelerated motion. In other words, motion at constant

acceleration.

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The following are the variables that will be used in this section:

vi = initial velocity(

m · s−1)

at t = 0s

vf = final velocity(

m · s−1)

at time t

∆x = displacement (m)

t = time (s)

∆t = time interval (s)

a = acceleration(

m · s−1)

(15.49)

vf = vi + at (15.50)

∆x =(vi + vf )

2t (15.51)

∆x = vit+1

2at2 (15.52)

v2f = v2i + 2a∆x (15.53)

The questions can vary a lot, but the following method for answering them will always work. Use this when

attempting a question that involves motion with constant acceleration. You need any three known quantities (vi,vf , ∆x, t or a) to be able to calculate the fourth one.

1. Read the question carefully to identify the quantities that are given. Write them down.

2. Identify the equation to use. Write it down!!!

3. Ensure that all the values are in the correct unit and fill them in your equation.

4. Calculate the answer and fill in its unit.

NOTE: Galileo Galilei of Pisa, Italy, was the first to determined the correct mathemat-

ical law for acceleration: the total distance covered, starting from rest, is proportional

to the square of the time. He also concluded that objects retain their velocity unless

a force – often friction – acts upon them, refuting the accepted Aristotelian hypothe-

sis that objects "naturally" slow down and stop unless a force acts upon them. This

principle was incorporated into Newton’s laws of motion (1st law).

1.115.1 Finding the Equations of Motion

The following does not form part of the syllabus and can be considered additional information.

Derivation of (15.50)

According to the definition of acceleration:

a =∆v

t(15.54)

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where ∆v is the change in velocity, i.e. ∆v = vf - vi. Thus we have

a =vf−vi

t

vf = vi + at(15.55)

Derivation of (15.51)

We have seen that displacement can be calculated from the area under a velocity vs. time graph. For uniformly

accelerated motion the most complicated velocity vs. time graph we can have is a straight line. Look at the graph

below - it represents an object with a starting velocity of vi, accelerating to a final velocity vf over a total time t.

To calculate the final displacement we must calculate the area under the graph - this is just the area of the

rectangle added to the area of the triangle. This portion of the graph has been shaded for clarity.

Area = 12b× h

= 12 t× (vf − vi)

= 12vf t− 1

2vit

(15.56)

Area = ℓ× b

= t× vi

= vit

(15.57)

Displacement = Area +Area

∆x = vit+12vf t− 1

2vit

∆x =(vi+vf )

2 t

(15.58)

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Derivation of (15.52)

This equation is simply derived by eliminating the final velocity vf in (15.51). Remembering from (15.50) that

vf = vi + at (15.59)

then (15.51) becomes

∆x = vi+vi+at2 t

= 2vit+at2

2

∆x = vit+12at

2

(15.60)

Derivation of (15.53)

This equation is just derived by eliminating the time variable in the above equation. From (15.50) we know

t =vf − vi

a(15.61)

Substituting this into (15.52) gives

∆x = vi

(

vf−vi

a

)

+ 12a

(

vf−vi

a

)2

=vivf

a− v2

i

a+ 1

2a(

v2

f−2vivf+v2

i

a2

)

=vivf

a− v2

i

a+

v2

f

2a − vivf

a+

v2

i

2a

2a∆x = −2v2i + v2f + v2i

v2f = v2i + 2a∆x

(15.62)

This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent

of the time variable.

Exercise 15.8: Equations of motion A racing car is travelling north. It

accelerates uniformly covering a distance of 725 m in 10 s. If it has an

initial velocity of 10 m·s−1, find its acceleration.

Solution to Exercise

Step 1. We are given:

vi = 10 m · s−1

∆x = 725 m

t = 10 s

a = ?

(15.63)

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Step 2. If you struggle to find the correct equation, find the quantity that is

not given and then look for an equation that has this quantity in it.

We can use equation (15.52)

∆x = vit+1

2at2 (15.64)

Step 3.

∆x = vit+12at

2

725 m =(

10m · s−1 × 10 s)

+ 12a× (10 s)

2

725 m− 100 m =(

50 s2)

a

a = 12, 5 m · s−2

(15.65)

Step 4. The racing car is accelerating at 12,5 m·s−2 north.

Exercise 15.9: Equations of motion A motorcycle, travelling east,

starts from rest, moves in a straight line with a constant acceleration and

covers a distance of 64 m in 4 s. Calculate

1. its acceleration

2. its final velocity

3. at what time the motorcycle had covered half the total distance

4. what distance the motorcycle had covered in half the total time.

Solution to Exercise

Step 1. We are given:

vi = 0m · s−1 (because the object starts from rest.)

∆x = 64m

t = 4s

a = ?

vf = ?

t = ?at half the distance∆x = 32 m.

∆x = ?at half the time t = 2 s.

(15.66)

All quantities are in SI units.

Step 2. We can use (15.52)

∆x = vit+1

2at2 (15.67)

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Step 3.

∆x = vit+12at

2

64 m =(

0m · s−1 × 4 s)

+ 12a× (4 s)

2

64 m =(

8 s2)

a

a = 8m · s−2 east

(15.68)

Step 4. We can use (15.53) - remember we now also know the acceleration

of the object.

vf = vi + at (15.69)

Step 5.

vf = vi + at

vf = 0m · s−1 +(

8m · s−2)

(4 s)

= 32m · s−1 east

(15.70)

Step 6. We can use (15.52):

∆x = vi +12at

2

32 m =(

0m · s−1)

t+ 12

(

8m · s−2)

(t)2

32 m = 0 +(

4m · s−2)

t2

8 s2 = t2

t = 2, 83 s

(15.71)

Step 7. Half the time is 2 s, thus we have vi, a and t - all in the correct units.

We can use (15.52) to get the distance:

∆x = vit+12at

2

= (0) (2) + 12 (8) (2)

2

= 16m east

(15.72)

Step 8. a. The acceleration is 8 m · s−2 east

b. The velocity is 32 m · s−1 east

c. The time at half the distance is 2, 83 sd. The distance at half the time is 16 m east

Equations of motion

1. A car starts off at 10 m·s−1 and accelerates at 1 m·s−2 for 10 s. What is its final velocity?

2. A train starts from rest, and accelerates at 1 m·s−2 for 10 s. How far does it move?

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3. A bus is going 30 m·s−1 and stops in 5 s. What is its stopping distance for this speed?

4. A racing car going at 20 m·s−1 stops in a distance of 20 m. What is its acceleration?

5. A ball has a uniform acceleration of 4 m·s−1. Assume the ball starts from rest. Determine the velocity and

displacement at the end of 10 s.

6. A motorcycle has a uniform acceleration of 4 m·s−1. Assume the motorcycle has an initial velocity of

20 m·s−1. Determine the velocity and displacement at the end of 12 s.

7. An aeroplane accelerates uniformly such that it goes from rest to 144 km·hr−1in 8 s. Calculate the acceler-

ation required and the total distance that it has traveled in this time.

Find the answers with the shortcodes:

(1.) l1Q (2.) l1U (3.) l1P (4.) l1E (5.) l1m (6.) l1y

(7.) l1V

1.116 Applications in the Real-World

(section shortcode: P10024 )

What we have learnt in this chapter can be directly applied to road safety. We can analyse the relationship

between speed and stopping distance. The following worked example illustrates this application.

Exercise 15.10: Stopping distance A truck is travelling at a constant

velocity of 10 m·s−1when the driver sees a child 50 m in front of him in

the road. He hits the brakes to stop the truck. The truck accelerates at

a rate of -1.25 m·s−2. His reaction time to hit the brakes is 0,5 seconds.

Will the truck hit the child?

Solution to Exercise

Step 1. It is useful to draw a timeline like this one:

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We need to know the following:

• What distance the driver covers before hitting the brakes.

• How long it takes the truck to stop after hitting the brakes.

• What total distance the truck covers to stop.

Step 2. Before the driver hits the brakes, the truck is travelling at constant

velocity. There is no acceleration and therefore the equations of

motion are not used. To find the distance traveled, we use:

v = Dt

10 = d0,5

d = 5 m

(15.73)

The truck covers 5 m before the driver hits the brakes.

Step 3. We have the following for the motion between B and C:

vi = 10m · s−1

vf = 0m · s−1

a = −1, 25m · s−2

t = ?

(15.74)

We can use (3.50)

vf = vi + at

0 = 10 + (−1, 25) t

−10 = −1, 25t

t = 8 s

(15.75)

Step 4. For the distance we can use (15.51) or (15.52). We will use (15.51):

∆x =(vi+vf )

2 t

∆x = 10+0s

(8)

∆x = 40 m

(15.76)

Step 5. The total distance that the truck covers is DAB + DBC = 5 + 40 =45 meters. The child is 50 meters ahead. The truck will not hit the

child.

1.117 Summary

(section shortcode: P10025 )

• A reference point is a point from where you take your measurements.

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• A frame of reference is a reference point with a set of directions.

• Your position is where you are located with respect to your reference point.

• The displacement of an object is how far it is from the reference point. It is the shortest distance between

the object and the reference point. It has magnitude and direction because it is a vector.

• The distance of an object is the length of the path travelled from the starting point to the end point. It has

magnitude only because it is a scalar.

• A vector is a physical quantity with magnitude and direction.

• A scalar is a physical quantity with magnitude only.

• Speed (s) is the distance covered (D) divided by the time taken (∆t):

s =D

∆t(15.77)

• Average velocity (v) is the displacement (∆x) divided by the time taken (∆t):

v =∆x

∆t(15.78)

• Instantaneous speed is the speed at a specific instant in time.

• Instantaneous velocity is the velocity at a specific instant in time.

• Acceleration (a) is the change in velocity (∆x) over a time interval (∆t):

a =∆v

∆t(15.79)

• The gradient of a position - time graph (x vs. t) give the velocity.

• The gradient of a velocity - time graph (v vs. t) give the acceleration.

• The area under a velocity - time graph (v vs. t) give the displacement.

• The area under an acceleration - time graph (a vs. t) gives the velocity.

• The graphs of motion are summarised in Figure 15.28.

• The equations of motion are used where constant acceleration takes place:

vf = vi + at

∆x =(vi+vf )

2 t

∆x = vit+12at

2

v2f = v2i + 2a∆x

(15.80)

1.118 End of Chapter Exercises: Motion in One Dimension

(section shortcode: P10026 )

1. Give one word/term for the following descriptions.

a. The shortest path from start to finish.

b. A physical quantity with magnitude and direction.

c. The quantity defined as a change in velocity over a time period.

d. The point from where you take measurements.

e. The distance covered in a time interval.

f. The velocity at a specific instant in time.

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2. Choose an item from column B that match the description in column A. Write down only the letter next to

the question number. You may use an item from column B more than once.

Column A Column B

a. The area under a velocity - time graph gradient

b. The gradient of a velocity - time graph area

c. The area under an acceleration - time graph velocity

d. The gradient of a displacement - time graph displacement

acceleration

slope

Table 15.5

3. Indicate whether the following statements are TRUE or FALSE. Write only ’true’ or ’false’. If the statement

is false, write down the correct statement.

a. A scalar is the displacement of an object over a time interval.

b. The position of an object is where it is located.

c. The sign of the velocity of an object tells us in which direction it is travelling.

d. The acceleration of an object is the change of its displacement over a period in time.

4. (SC 2003/11) A body accelerates uniformly from rest for t0 seconds after which it continues with a constant

velocity. Which graph is the correct representation of the body’s motion?

continued on next page

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(a) (b) (c) (d)

Table 15.6

5. (SC 2003/11) The velocity-time graphs of two cars are represented by P and Q as shown

The difference in the distance travelled by the two cars (in m) after 4 s is ...

a. 12

b. 6

c. 2

d. 0

6. (IEB 2005/11 HG) The graph that follows shows how the speed of an athlete varies with time as he sprints

for 100 m.

Which of the following equations can be used to correctly determine the time t for which he accelerates?

a. 100 = (10) (11)− 12 (10) t

b. 100 = (10) (11) + 12 (10) t

c. 100 = 10t+ 12 (10) t

2

d. 100 = 12 (0) t+

12 (10) t

2

7. (SC 2002/03 HG1) In which one of the following cases will the distance covered and the magnitude of the

displacement be the same?

a. A girl climbs a spiral staircase.

b. An athlete completes one lap in a race.

c. A raindrop falls in still air.

d. A passenger in a train travels from Cape Town to Johannesburg.

8. (SC 2003/11) A car, travelling at constant velocity, passes a stationary motor cycle at a traffic light. As

the car overtakes the motorcycle, the motorcycle accelerates uniformly from rest for 10 s. The following

displacement-time graph represents the motions of both vehicles from the traffic light onwards.

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a. Use the graph to find the magnitude of the constant velocity of the car.

b. Use the information from the graph to show by means of calculation that the magnitude of the acceler-

ation of the motorcycle, for the first 10 s of its motion is 7,5 m·s−2.

c. Calculate how long (in seconds) it will take the motorcycle to catch up with the car (point X on the time

axis).

d. How far behind the motorcycle will the car be after 15 seconds?

9. (IEB 2005/11 HG) Which of the following statements is true of a body that accelerates uniformly?

a. Its rate of change of position with time remains constant.

b. Its position changes by the same amount in equal time intervals.

c. Its velocity increases by increasing amounts in equal time intervals.

d. Its rate of change of velocity with time remains constant.

10. (IEB 2003/11 HG1) The velocity-time graph for a car moving along a straight horizontal road is shown below.

Which of the following expressions gives the magnitude of the average velocity of the car?

a. AreaAt

b. AreaA + AreaBt

c. AreaBt

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d. AreaA − AreaBt

11. (SC 2002/11 SG) A car is driven at 25 m·s−1 in a municipal area. When the driver sees a traffic officer at a

speed trap, he realises he is travelling too fast. He immediately applies the brakes of the car while still 100

m away from the speed trap.

a. Calculate the magnitude of the minimum acceleration which the car must have to avoid exceeding the

speed limit, if the municipal speed limit is 16.6 m·s−1.

b. Calculate the time from the instant the driver applied the brakes until he reaches the speed trap. Assume

that the car’s velocity, when reaching the trap, is 16.6 m·s−1.

12. A traffic officer is watching his speed trap equipment at the bottom of a valley. He can see cars as they

enter the valley 1 km to his left until they leave the valley 1 km to his right. Nelson is recording the times

of cars entering and leaving the valley for a school project. Nelson notices a white Toyota enter the valley

at 11:01:30 and leave the valley at 11:02:42. Afterwards, Nelson hears that the traffic officer recorded the

Toyota doing 140 km·hr−1.

a. What was the time interval (∆t) for the Toyota to travel through the valley?

b. What was the average speed of the Toyota?

c. Convert this speed to km·hr−1.

d. Discuss whether the Toyota could have been travelling at 140km·hr−1 at the bottom of the valley.

e. Discuss the differences between the instantaneous speed (as measured by the speed trap) and average

speed (as measured by Nelson).

13. (IEB 2003/11HG) A velocity-time graph for a ball rolling along a track is shown below. The graph has been

divided up into 3 sections, A, B and C for easy reference. (Disregard any effects of friction.)

a. Use the graph to determine the following:

i. the speed 5 s after the start

ii. the distance travelled in Section A

iii. the acceleration in Section C

b. At time t1 the velocity-time graph intersects the time axis. Use an appropriate equation of motion to

calculate the value of time t1 (in s).

c. Sketch a displacement-time graph for the motion of the ball for these 12 s. (You do not need to calculate

the actual values of the displacement for each time interval, but do pay attention to the general shape

of this graph during each time interval.)

14. In towns and cities, the speed limit is 60 km·hr−1. The length of the average car is 3.5 m, and the width of

the average car is 2 m. In order to cross the road, you need to be able to walk further than the width of a

car, before that car reaches you. To cross safely, you should be able to walk at least 2 m further than the

width of the car (4 m in total), before the car reaches you.

a. If your walking speed is 4 km·hr−1, what is your walking speed in m·s−1?

b. How long does it take you to walk a distance equal to the width of the average car?

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c. What is the speed in m·s−1 of a car travelling at the speed limit in a town?

d. How many metres does a car travelling at the speed limit travel, in the same time that it takes you to

walk a distance equal to the width of car?

e. Why is the answer to the previous question important?

f. If you see a car driving toward you, and it is 28 m away (the same as the length of 8 cars), is it safe to

walk across the road?

g. How far away must a car be, before you think it might be safe to cross? How many car-lengths is this

distance?

15. A bus on a straight road starts from rest at a bus stop and accelerates at 2 m·s−2 until it reaches a speed

of 20 m·s−1. Then the bus travels for 20 s at a constant speed until the driver sees the next bus stop in the

distance. The driver applies the brakes, stopping the bus in a uniform manner in 5 s.

a. How long does the bus take to travel from the first bus stop to the second bus stop?

b. What is the average velocity of the bus during the trip?

Find the answers with the shortcodes:

(1.) lrr (2.) lrY (3.) lrg (4.) lr4 (5.) lr2 (6.) lrT

(7.) lrb (8.) lrj (9.) lrD (10.) lrW (11.) lrZ (12.) lrB

(13.) lrK (14.) lrk (15.) lr0

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Transverse Pulses

1.119 Introduction

(section shortcode: P10027 )

This chapter forms the basis of the discussion into mechanical waves. Waves are all around us, even though

most of us are not aware of it. The most common waves are waves in the sea, but waves can be created in any

container of water, ranging from an ocean to a tea-cup. Waves do not only occur in water, they occur in any kind

of medium. Earthquakes generate waves that travel through the rock of the Earth. When your friend speaks to

you he produces sound waves that travel through the air to your ears. Light is made up of electromagnetic waves.

A wave is simply moving energy.

1.120 What is a medium?

(section shortcode: P10028 )

In this chapter, as well as in the following chapters, we will speak about waves moving in a medium. A medium is

just the substance or material through which waves move. In other words the medium carries the wave from one

place to another. The medium does not create the wave and the medium is not the wave. Therefore the medium

does not travel with the wave as the wave propagates through it. Air is a medium for sound waves, water is a

medium for water waves and rock is a medium for earthquakes (which are also a type of wave). Air, water and

rock are therefore examples of media (media is the plural of medium).

Definition: Medium

A medium is the substance or material in which a wave will move.

In each medium, the atoms that make up the medium are moved temporarily from their rest position. In order for

a wave to travel, the different parts of the medium must be able to interact with each other.

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1.121. WHAT IS A PULSE? CHAPTER 1. TRANSVERSE PULSES

1.121 What is a pulse?

(section shortcode: P10029 )

1.121.1 Investigation : Observation of Pulses

Take a heavy rope. Have two people hold the rope stretched out horizontally. Flick the rope at one end only once.

What happens to the disturbance that you created in the rope? Does it stay at the place where it was created or

does it move down the length of the rope?

In the activity, we created a pulse. A pulse is a single disturbance that moves through a medium. In a transverse

pulse the displacement of the medium is perpendicular to the direction of motion of the pulse. Figure 16.2 shows

an example of a transverse pulse. In the activity, the rope or spring was held horizontally and the pulse moved

the rope up and down. This was an example of a transverse pulse.

Definition: Pulse

A pulse is a single disturbance that moves through a medium.

Definition: Transverse Pulse

A pulse where all of the particles disturbed by the pulse move perpendicular (at a right

angle) to the direction in which the pulse is moving.

1.121.2 Pulse Length and Amplitude

The amplitude of a pulse is a measurement of how far the medium is displaced momentarily from a position of

rest. The pulse length is a measurement of how long the pulse is. Both these quantities are shown in Figure 16.2.

Definition: Amplitude

The amplitude of a pulse is a measurement of how far the medium is displaced from rest.

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CHAPTER 1. TRANSVERSE PULSES 1.121. WHAT IS A PULSE?

Figure 16.2: Example of a transverse pulse

The position of rest is the position the medium would be in if it were undisturbed. This is also called the equilibrium

position. Sometimes people will use rest and sometimes equilibrium but they will also use to the two in the same

discussion to mean the same thing.

Investigation : Pulse Length and Amplitude

The graphs below show the positions of a pulse at different times.

Use your ruler to measure the lengths of a and p. Fill your answers in the table.

Time a p

t = 0 s

t = 1 s

t = 2 s

t = 3 s

Table 16.1

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1.121. WHAT IS A PULSE? CHAPTER 1. TRANSVERSE PULSES

What do you notice about the values of a and p?

In the activity, we found that the values for how high the pulse (a) is and how wide the pulse (p) is the same at

different times. Pulse length and amplitude are two important quantities of a pulse.

1.121.3 Pulse Speed

Definition: Pulse Speed

Pulse speed is the distance a pulse travels per unit time.

In Motion in one dimension (Chapter 15) we saw that speed was defined as the distance traveled per unit time.

We can use the same definition of speed to calculate how fast a pulse travels. If the pulse travels a distance D in

a time t, then the pulse speed v is:

v =D

t(16.1)

Exercise 16.1: Pulse Speed A pulse covers a distance of 2 m in 4 son a heavy rope. Calculate the pulse speed.

Solution to Exercise

Step 1. We are given:

• the distance travelled by the pulse: D = 2m• the time taken to travel 2 m: t = 4 s

We are required to calculate the speed of the pulse.

Step 2. We can use:

v =D

t(16.2)

to calculate the speed of the pulse.

Step 3.

v = Dt

= 2 m4 s

= 0, 5m · s−1

(16.3)

Step 4. The pulse speed is 0, 5 m · s−1.

TIP: The pulse speed depends on the properties of the medium and not on the ampli-

tude or pulse length of the pulse.

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CHAPTER 1. TRANSVERSE PULSES 1.122. SUPERPOSITION OF PULSES

Pulse Speed

1. A pulse covers a distance of 5 m in 15 s. Calculate the speed of the pulse.

2. A pulse has a speed of 5 cm · s−1. How far does it travel in 2, 5 s?

3. A pulse has a speed of 0, 5 m · s−1. How long does it take to cover a distance of 25 cm?

4. How long will it take a pulse moving at 0, 25 m · s−1 to travel a distance of 20 m?

5. The diagram shows two pulses in the same medium. Which has the higher speed? Explain your answer.

Find the answers with the shortcodes:

(1.) l1f (2.) l1G (3.) l17 (4.) l1A (5.) l1o

1.122 Superposition of Pulses

(section shortcode: P10030 )

Two or more pulses can pass through the same medium at that same time in the same place. When they do they

interact with each other to form a different disturbance at that point. The resulting pulse is obtained by using the

principle of superposition. The principle of superposition states that the effect of the different pulses is the sum

of their individual effects. After pulses pass through each other, each pulse continues along its original direction

of travel, and their original amplitudes remain unchanged.

Constructive interference takes place when two pulses meet each other to create a larger pulse. The amplitude

of the resulting pulse is the sum of the amplitudes of the two initial pulses. This is shown in Figure 16.5.

Definition: Constructive interference

Constructive interference is when two pulses meet, resulting in a bigger pulse.

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1.122. SUPERPOSITION OF PULSES CHAPTER 1. TRANSVERSE PULSES

Figure 16.5: Superposition of two pulses: constructive interference.

Destructive interference takes place when two pulses meet and cancel each other. The amplitude of the resulting

pulse is the sum of the amplitudes of the two initial pulses, but the one amplitude will be a negative number. This

is shown in Figure 16.6. In general, amplitudes of individual pulses add together to give the amplitude of the

resultant pulse.

Definition: Destructive interference

Destructive interference is when two pulses meet, resulting in a smaller pulse.

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CHAPTER 1. TRANSVERSE PULSES 1.122. SUPERPOSITION OF PULSES

Figure 16.6: Superposition of two pulses. The left-hand series of images demonstrates destructive interference,

since the pulses cancel each other. The right-hand series of images demonstrate a partial cancelation of two

pulses, as their amplitudes are not the same in magnitude.

Exercise 16.2: Superposition of Pulses The two pulses shown below

approach each other at 1 m · s−1. Draw what the waveform would look

like after 1 s, 2 s and 5 s.

Solution to Exercise

Step 1. After 1 s, pulse A has moved 1 m to the right and pulse B has moved

1 m to the left.

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1.122. SUPERPOSITION OF PULSES CHAPTER 1. TRANSVERSE PULSES

Step 2. After 1 s more, pulse A has moved 1 m to the right and pulse B has

moved 1 m to the left.

Step 3. After 5 s, pulse A has moved 5 m to the right and pulse B has moved

5 m to the left.

TIP: The idea of superposition is one that occurs often in physics. You will see much,

much more of superposition!

1.122.1 Experiment: Constructive and destructive interference

Aim: To demonstrate constructive and destructive interference

Apparatus: Ripple tank apparatus

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CHAPTER 1. TRANSVERSE PULSES 1.122. SUPERPOSITION OF PULSES

Method:

1. Set up the ripple tank

2. Produce a single pulse and observe what happens

3. Produce two pulses simultaneously and observe what happens

4. Produce two pulses at slightly different times and observe what happens

Results and conclusion: You should observe that when you produce two pulses simultaneously you see them

interfere constructively and when you produce two pulses at slightly different times you see them interfere de-

structively.

1.122.2 Problems Involving Superposition of Pulses

1. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s, 4 s and 5 s. Each pulse is travelling

at 1 m · s−1. Each block represents 1 m. The pulses are shown as thick black lines and the undisplaced

medium as dashed lines.

2. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s, 4 s and 5 s. Each pulse is travelling

at 1 m · s−1. Each block represents 1 m. The pulses are shown as thick black lines and the undisplaced

medium as dashed lines.

3. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s, 4 s and 5 s. Each pulse is travelling

at 1 m · s−1. Each block represents 1 m. The pulses are shown as thick black lines and the undisplaced

medium as dashed lines.

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1.122. SUPERPOSITION OF PULSES CHAPTER 1. TRANSVERSE PULSES

4. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s, 4 s and 5 s. Each pulse is travelling

at 1 m · s−1. Each block represents 1 m. The pulses are shown as thick black lines and the undisplaced

medium as dashed lines.

5. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s, 4 s and 5 s. Each pulse is travelling

at 1 m · s−1. Each block represents 1 m. The pulses are shown as thick black lines and the undisplaced

medium as dashed lines.

6. For the following pulse, draw the resulting wave forms after 1 s, 2 s, 3 s, 4 s and 5 s. Each pulse is travelling

at 1 m · s−1. Each block represents 1 m. The pulses are shown as thick black lines and the undisplaced

medium as dashed lines.

7. What is superposition of waves?

8. What is constructive interference?

9. What is destructive interference?

The following presentation provides a summary of the work covered in this chapter. Although the presentation is

titled waves, the presentation covers pulses only.

(Presentation: P10031 )

Find the answers with the shortcodes:

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CHAPTER 1. TRANSVERSE PULSES1.123. GRAPHS OF POSITION AND VELOCITY (NOT IN CAPS - INCLUDED FOR COMPLETENESS)

(1.) l1M (2.) l1e (3.) l1t (4.) l1z (5.) l1u (6.) l1J

(7.) l1S (8.) l1h (9.) lrg

1.123 Graphs of Position and Velocity (Not in CAPS - Included for Com-

pleteness)

(section shortcode: P10032 )

When a pulse moves through a medium, there are two different motions: the motion of the particles of the medium

and the motion of the pulse. These two motions are at right angles to each other when the pulse is transverse.

Each motion will be discussed.

Consider the situation shown in Figure 16.20. The dot represents one particle of the medium. We see that as the

pulse moves to the right the particle only moves up and down.

1.123.1 Motion of a Particle of the Medium

First we consider the motion of a particle of the medium when a pulse moves through the medium. For the

explanation we will zoom into the medium so that we are looking at the atoms of the medium. These atoms are

connected to each other as shown in Figure 16.19.

Figure 16.19: Particles in a medium.

When a pulse moves through the medium, the particles in the medium only move up and down. We can see this

in Figure 16.20 which shows the motion of a single particle as a pulse moves through the medium.

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1.123. GRAPHS OF POSITION AND VELOCITY (NOT IN CAPS - INCLUDED FOR COMPLETENESS)CHAPTER 1. TRANSVERSE PULSES

Figure 16.20: Positions of a pulse in a rope at different times. The pulse moves to the right as shown by the

arrow. You can also see the motion of a point in the medium through which the pulse is travelling. Each block is

1 cm.

TIP: A particle in the medium only moves up and down when a transverse pulse moves

through the medium. The pulse moves from left to right (or right to left). The motion of

the particle is perpendicular to the motion of a transverse pulse.

If you consider the motion of the particle as a function of time, you can draw a graph of position vs. time and

velocity vs. time.

Investigation : Drawing a position-time graph

1. Study Figure 16.20 and complete the following table:

time (s) 0 1 2 3 4 5 6 7 8 9

position (cm)

Table 16.2

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CHAPTER 1. TRANSVERSE PULSES1.123. GRAPHS OF POSITION AND VELOCITY (NOT IN CAPS - INCLUDED FOR COMPLETENESS)

2. Use your table to draw a graph of position vs. time for a particle in a medium.

The position vs. time graph for a particle in a medium when a pulse passes through the medium is shown in

Figure 16.21

Figure 16.21: Position against Time graph of a particle in the medium through which a transverse pulse is

travelling.

Investigation : Drawing a velocity-time graph

1. Study Figure 16.21 and complete the following table:

time (s) 0 1 2 3 4 5 6 7 8 9

velocity (cm.s−1)

Table 16.3

2. Use your table to draw a graph of velocity vs time for a particle in a medium.

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The velocity vs. time graph far a particle in a medium when a pulse passes through the medium is shown in

Figure 16.22.

Figure 16.22: Velocity against Time graph of a particle in the medium through which a transverse pulse is

travelling.

1.123.2 Motion of the Pulse

The motion of the pulse is much simpler than the motion of a particle in the medium.

TIP: A point on a transverse pulse, eg. the peak, only moves in the direction of the

motion of the pulse.

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CHAPTER 1. TRANSVERSE PULSES1.123. GRAPHS OF POSITION AND VELOCITY (NOT IN CAPS - INCLUDED FOR COMPLETENESS)

Exercise 16.3: Transverse pulse through a medium

Figure 16.23: Position of the peak of a pulse at different times (since we know

the shape of the pulse does not change we can look at only one point on the

pulse to keep track of its position, the peak for example). The pulse moves to

the right as shown by the arrow. Each square is 0, 5 cm.

Given the series of snapshots of a transverse pulse moving through a

medium, depicted in Figure 16.23, do the following:

• draw up a table of time, position and velocity,

• plot a position vs. time graph,

• plot a velocity vs. time graph.

Solution to Exercise

Step 1. Figure 16.23 shows the motion of a pulse through a medium and

a dot to indicate the same position on the pulse. If we follow the

dot, we can draw a graph of position vs time for a pulse. At t = 0 sthe dot is at 0 cm. At t = 1 s the dot is 1 cm away from its original

postion. At t = 2 s the dot is 2 cm away from its original postion,

and so on.

Step 2.

time (s) 0 1 2 3 4 5 6 7 8 9

position (cm) 0 1 2 3 4 5 6 7 8 9

velocity (cm.s−1) 1 1 1 1 1 1 1 1 1 1

Table 16.4

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Step 3.

Step 4.

Travelling Pulse

1. A pulse is passed through a rope and the following pictures were obtained for each time interval:

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CHAPTER 1. TRANSVERSE PULSES1.124. TRANSMISSION AND REFLECTION OF A PULSE AT A BOUNDARY

a. Complete the following table for a particle in the medium:

time (s) 0,00 0,25 0,50 0,75 1,00 1,25 1,50 1,75 2,00

position (mm)

velocity (mm.s−1)

Table 16.5

b. Draw a position vs. time graph for the motion of the particle at 3 cm.

c. Draw a velocity vs. time graph for the motion of the particle at 3 cm.

d. Draw a position vs. time graph for the motion of the pulse through the rope.

e. Draw a velocity vs. time graph for the motion of the pulse through the rope.

Find the answers with the shortcodes:

(1.) l1s

1.124 Transmission and Reflection of a Pulse at a Boundary

(section shortcode: P10033 )

What happens when a pulse travelling in one medium finds that medium is joined to another?

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1.124. TRANSMISSION AND REFLECTION OF A PULSE AT A BOUNDARYCHAPTER 1. TRANSVERSE PULSES

1.124.1 Investigation : Two ropes

Find two different ropes and tie both ropes together. Hold the joined ropes horizontally and create a pulse by

flicking the rope up and down. What happens to the pulse when it encounters the join?

When a pulse is transmitted from one medium to another, like from a thin rope to a thicker one, the nature of the

pulse will change where it meets the boundary of the two media (i.e. where the two ropes are joined). Part of the

pulse will be reflected and part of it will be transmitted. Figure 16.27 shows the general case of a pulse meeting a

boundary. The incident pulse is the one that arrives at the boundary. The reflected pulse is the one that moves

back, away from the boundary. The transmitted pulse is the one that moves into the new medium, away from

the boundary. The speed of the pulse depends on the mass of the rope; the pulse is faster in the thinner rope

and slower in the thick rope. When the speed of the pulse increases, the pulse length will increase. If the speed

decreases, the pulse length will decrease.

Figure 16.27: Reflection and transmission of a pulse at the boundary between two media.

Consider a pulse moving from a thin rope to a thick rope. As the pulse crosses the boundary, the speed of

the pulse will decrease as it moves into the thicker rope. The pulse will move slower, so the pulse length will

decrease. The pulse will be reflected and inverted in the thin rope. The reflected pulse will have the same length

and speed but will have a smaller amplitude. This is illustrated in Figure 16.28.

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CHAPTER 1. TRANSVERSE PULSES1.124. TRANSMISSION AND REFLECTION OF A PULSE AT A BOUNDARY

Figure 16.28: Reflection and transmission of a pulse at the boundary between two media.

When a pulse moves from a thick rope to a thin rope, the opposite will happen. As the pulse crosses the boundary,

the speed of the pulse will increase as it moves into the thinner rope. The pulse in the thin rope will move faster,

so the pulse length will increase. The pulse will be reflected but not inverted in the thick rope. The reflected pulse

will have the same length and speed but will have a smaller amplitude. This is illustrated in Figure 16.29

Figure 16.29: Reflection and transmission of a pulse at the boundary between two media.

1.124.2 Pulses at a Boundary I

1. Fill in the blanks or select the correct answer: A pulse in a heavy rope is traveling towards the boundary with

a thin piece of string.

a. The reflected pulse in the heavy rope will/will not be inverted because .

b. The speed of the transmitted pulse will be greater than/less than/the same as the speed of the

incident pulse.

c. The speed of the reflected pulse will be greater than/less than/the same as the speed of the incident

pulse.

d. The pulse length of the transmitted pulse will be greater than/less than/the same as the pulse length

of the incident pulse.

e. The frequency of the transmitted pulse will be greater than/less than/the same as the frequency of

the incident pulse.

2. A pulse in a light string is traveling towards the boundary with a heavy rope.

a. The reflected pulse in the light rope will/will not be inverted because .

b. The speed of the transmitted pulse will be greater than/less than/the same as the speed of the

incident pulse.

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1.125. REFLECTION OF A PULSE FROM FIXED AND FREE ENDS (NOT IN CAPS - INCLUDED FOR

COMPLETENESS) CHAPTER 1. TRANSVERSE PULSES

c. The speed of the reflected pulse will be greater than/less than/the same as the speed of the incident

pulse.

d. The pulse length of the transmitted pulse will be greater than/less than/the same as the pulse length

of the incident pulse.

Find the answers with the shortcodes:

(1.) l1H (2.) l16

1.125 Reflection of a Pulse from Fixed and Free Ends (not in CAPS -

included for completeness)

(section shortcode: P10034 )

Let us now consider what happens to a pulse when it reaches the end of a medium. The medium can be fixed,

like a rope tied to a wall, or it can be free, like a rope tied loosely to a pole.

1.125.1 Reflection of a Pulse from a Fixed End

Investigation : Reflection of a Pulse from a Fixed End

Tie a rope to a wall or some other object that cannot move. Create a pulse in the rope by flicking one end up and

down. Observe what happens to the pulse when it reaches the wall.

Figure 16.30: Reflection of a pulse from a fixed end.

When the end of the medium is fixed, for example a rope tied to a wall, a pulse reflects from the fixed end, but

the pulse is inverted (i.e. it is upside-down). This is shown in Figure 16.30.

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CHAPTER 1. TRANSVERSE PULSES

1.125. REFLECTION OF A PULSE FROM FIXED AND FREE ENDS (NOT IN CAPS - INCLUDED FOR

COMPLETENESS)

1.125.2 Reflection of a Pulse from a Free End

Investigation : Reflection of a Pulse from a Free End

Tie a rope to a pole in such a way that the rope can move up and down the pole. Create a pulse in the rope by

flicking one end up and down. Observe what happens to the pulse when it reaches the pole.

When the end of the medium is free, for example a rope tied loosely to a pole, a pulse reflects from the free end,

but the pulse is not inverted. This is shown in Figure 16.31. We draw the free end as a ring around the pole. The

ring will move up and down the pole, while the pulse is reflected away from the pole.

Figure 16.31: Reflection of a pulse from a free end.

TIP: The fixed and free ends that were discussed in this section are examples of

boundary conditions. You will see more of boundary conditions as you progress in the

Physics syllabus.

Pulses at a Boundary II

1. A rope is tied to a tree and a single pulse is generated. What happens to the pulse as it reaches the tree?

Draw a diagram to explain what happens.

2. A rope is tied to a ring that is loosely fitted around a pole. A single pulse is sent along the rope. What will

happen to the pulse as it reaches the pole? Draw a diagram to explain your answer.

The following simulation will help you understand the previous examples. Choose pulse from the options (either

manual, oscillate or pulse). Then click on pulse and see what happens. Change from a fixed to a free end and

see what happens. Try varying the width, amplitude, damping and tension.

Phet simulation for transverse pulses (Simulation: P10035 )

Find the answers with the shortcodes:

(1.) l1F (2.) l1L

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1.126. SUMMARY CHAPTER 1. TRANSVERSE PULSES

1.126 Summary

(section shortcode: P10036 )

• A medium is the substance or material in which a wave will move

• A pulse is a single disturbance that moves through a medium

• The amplitude of a pules is a measurement of how far the medium is displaced from rest

• Pulse speed is the distance a pulse travels per unit time

• Constructive interference is when two pulses meet and result in a bigger pulse

• Destructive interference is when two pulses meet and and result in a smaller pulse

• We can draw graphs to show the motion of a particle in the medium or to show the motion of a pulse through

the medium

• When a pulse moves from a thin rope to a thick rope, the speed and pulse length decrease. The pulse will

be reflected and inverted in the thin rope. The reflected pulse has the same length and speed, but a different

amplitude

• When a pulse moves from a thick rope to a thin rope, the speed and pulse length increase. The pulse will

be reflected in the thick rope. The reflected pulse has the same length and speed, but a different amplitude

• A pulse reaching a free end will be reflected but not inverted. A pulse reaching a fixed end will be reflected

and inverted

1.127 Exercises - Transverse Pulses

(section shortcode: P10037 )

1. A heavy rope is flicked upwards, creating a single pulse in the rope. Make a drawing of the rope and indicate

the following in your drawing:

a. The direction of motion of the pulse

b. Amplitude

c. Pulse length

d. Position of rest

2. A pulse has a speed of 2, 5 m · s−1. How far will it have travelled in 6 s?

3. A pulse covers a distance of 75 cm in 2, 5 s. What is the speed of the pulse?

4. How long does it take a pulse to cover a distance of 200 mm if its speed is 4 m · s−1?

5. The following position-time graph for a pulse in a slinky spring is given. Draw an accurate sketch graph of

the velocity of the pulse against time.

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CHAPTER 1. TRANSVERSE PULSES 1.127. EXERCISES - TRANSVERSE PULSES

6. The following velocity-time graph for a particle in a medium is given. Draw an accurate sketch graph of the

position of the particle vs. time.

7. Describe what happens to a pulse in a slinky spring when:

a. the slinky spring is tied to a wall.

b. the slinky spring is loose, i.e. not tied to a wall.

(Draw diagrams to explain your answers.)

8. The following diagrams each show two approaching pulses. Redraw the diagrams to show what type of

interference takes place, and label the type of interference.

a.

b.

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9. Two pulses, A and B, of identical shape and amplitude are simultaneously generated in two identical wires

of equal mass and length. Wire A is, however, pulled tighter than wire B. Which pulse will arrive at the other

end first, or will they both arrive at the same time?

Find the answers with the shortcodes:

(1.) lrl (2.) lri (3.) lr3 (4.) lrO (5.) lrc (6.) lrx

(7.) lra (8.) lrC (9.) lr1

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Transverse Waves

1.128 Introduction

(section shortcode: P10038 )

Waves occur frequently in nature. The most obvious examples are waves in water, on a dam, in the ocean, or in

a bucket. We are most interested in the properties that waves have. All waves have the same properties, so if we

study waves in water, then we can transfer our knowledge to predict how other examples of waves will behave.

1.129 What is a transverse wave?

(section shortcode: P10039 )

We have studied pulses in Transverse Pulses (Chapter 16), and know that a pulse is a single disturbance that

travels through a medium. A wave is a periodic, continuous disturbance that consists of a train or succession of

pulses.

Definition: Wave

A wave is a periodic, continuous disturbance that consists of a train of pulses.

Definition: Transverse wave

A transverse wave is a wave where the movement of the particles of the medium is per-

pendicular (at a right angle) to the direction of propagation of the wave.

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1.129. WHAT IS A TRANSVERSE WAVE? CHAPTER 1. TRANSVERSE WAVES

1.129.1 Investigation : Transverse Waves

Take a rope or slinky spring. Have two people hold the rope or spring stretched out horizontally. Flick the one

end of the rope up and down continuously to create a train of pulses.

1. Describe what happens to the rope.

2. Draw a diagram of what the rope looks like while the pulses travel along it.

3. In which direction do the pulses travel?

4. Tie a ribbon to the middle of the rope. This indicates a particle in the rope.

5. Flick the rope continuously. Watch the ribbon carefully as the pulses travel through the rope. What happens

to the ribbon?

6. Draw a picture to show the motion of the ribbon. Draw the ribbon as a dot and use arrows to indicate how it

moves.

In the Activity, you have created waves. The medium through which these waves propagated was the rope, which

is obviously made up of a very large number of particles (atoms). From the activity, you would have noticed that

the wave travelled from left to right, but the particles (the ribbon) moved only up and down.

Figure 17.3: A transverse wave, showing the direction of motion of the wave perpendicular to the direction in

which the particles move.

When the particles of a medium move at right angles to the direction of propagation of a wave, the wave is called

transverse. For waves, there is no net displacement of the particles (they return to their equilibrium position), but

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there is a net displacement of the wave. There are thus two different motions: the motion of the particles of the

medium and the motion of the wave.

The following simulation will help you understand more about waves. Select the oscillate option and then observe

what happens.

Phet simulation for Transverse Waves (Simulation: P10040 )

1.129.2 Peaks and Troughs

Waves have moving peaks (or crests) and troughs. A peak is the highest point the medium rises to and a trough

is the lowest point the medium sinks to.

Peaks and troughs on a transverse wave are shown in Figure 17.5.

Figure 17.5: Peaks and troughs in a transverse wave.

Definition: Peaks and troughs

A peak is a point on the wave where the displacement of the medium is at a maximum.

A point on the wave is a trough if the displacement of the medium at that point is at a

minimum.

1.129.3 Amplitude and Wavelength

There are a few properties that we saw with pulses that also apply to waves. These are amplitude and wavelength

(we called this pulse length).

Definition: Amplitude

The amplitude is the maximum displacement of a particle from its equilibrium position.

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Investigation : Amplitude

Fill in the table below by measuring the distance between the equilibrium and each peak and troughs in the wave

above. Use your ruler to measure the distances.

Peak/Trough Measurement (cm)

a

b

c

d

e

f

Table 17.1

1. What can you say about your results?

2. Are the distances between the equilibrium position and each peak equal?

3. Are the distances between the equilibrium position and each trough equal?

4. Is the distance between the equilibrium position and peak equal to the distance between equilibrium and

trough?

As we have seen in the activity on amplitude, the distance between the peak and the equilibrium position is equal

to the distance between the trough and the equilibrium position. This distance is known as the amplitude of the

wave, and is the characteristic height of wave, above or below the equilibrium position. Normally the symbol A is

used to represent the amplitude of a wave. The SI unit of amplitude is the metre (m).

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Exercise 17.1: Amplitude of Sea Waves If the peak of a wave mea-

sures 2 m above the still water mark in the harbour, what is the amplitude

of the wave?

Solution to Exercise

Step 1. The definition of the amplitude is the height of a peak above the

equilibrium position. The still water mark is the height of the water

at equilibrium and the peak is 2 m above this, so the amplitude is

2 m.

Investigation : Wavelength

Fill in the table below by measuring the distance between peaks and troughs in the wave above.

Distance(cm)

a

b

c

d

Table 17.2

1. What can you say about your results?

2. Are the distances between peaks equal?

3. Are the distances between troughs equal?

4. Is the distance between peaks equal to the distance between troughs?

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As we have seen in the activity on wavelength, the distance between two adjacent peaks is the same no matter

which two adjacent peaks you choose. There is a fixed distance between the peaks. Similarly, we have seen that

there is a fixed distance between the troughs, no matter which two troughs you look at. More importantly, the

distance between two adjacent peaks is the same as the distance between two adjacent troughs. This distance

is called the wavelength of the wave.

The symbol for the wavelength is λ (the Greek letter lambda) and wavelength is measured in metres (m).

Exercise 17.2: Wavelength The total distance between 4 consecutive

peaks of a transverse wave is 6 m. What is the wavelength of the wave?

Solution to Exercise

Step 1.

Step 2. From the sketch we see that 4 consecutive peaks is equivalent to 3

wavelengths.

Step 3. Therefore, the wavelength of the wave is:

3λ = 6m

λ = 6 m3

= 2m

(17.1)

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1.129.4 Points in Phase

Investigation : Points in Phase

Fill in the table by measuring the distance between the indicated points.

Points Distance (cm)

A to F

B to G

C to H

D to I

E to J

Table 17.3

What do you find?

In the activity the distance between the indicated points was the same. These points are then said to be in phase.

Two points in phase are separate by an integer (0,1,2,3,...) number of complete wave cycles. They do not have

to be peaks or troughs, but they must be separated by a complete number of wavelengths.

We then have an alternate definition of the wavelength as the distance between any two adjacent points which

are in phase.

Definition: Wavelength of wave

The wavelength of a wave is the distance between any two adjacent points that are in

phase.

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Points that are not in phase, those that are not separated by a complete number of wavelengths, are called out

of phase. Examples of points like these would be A and C, or D and E, or B and H in the Activity.

1.129.5 Period and Frequency

Imagine you are sitting next to a pond and you watch the waves going past you. First one peak arrives, then a

trough, and then another peak. Suppose you measure the time taken between one peak arriving and then the

next. This time will be the same for any two successive peaks passing you. We call this time the period, and it is

a characteristic of the wave.

The symbol T is used to represent the period. The period is measured in seconds (s).

Definition: Period (T)

The period (T) is the time taken for two successive peaks (or troughs) to pass a fixed

point.

Imagine the pond again. Just as a peak passes you, you start your stopwatch and count each peak going past.

After 1 second you stop the clock and stop counting. The number of peaks that you have counted in the 1 second

is the frequency of the wave.

Definition: Frequency

The frequency is the number of successive peaks (or troughs) passing a given point in 1

second.

The frequency and the period are related to each other. As the period is the time taken for 1 peak to pass, then

the number of peaks passing the point in 1 second is 1T

. But this is the frequency. So

f =1

T(17.2)

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or alternatively,

T =1

f(17.3)

For example, if the time between two consecutive peaks passing a fixed point is 12 s, then the period of the wave

is 12 s. Therefore, the frequency of the wave is:

f = 1T

= 11

2s

= 2 s−1

(17.4)

The unit of frequency is the Hertz (Hz) or s−1.

Exercise 17.3: Period and Frequency What is the period of a wave of

frequency 10 Hz?

Solution to Exercise

Step 1. We are required to calculate the period of a 10 Hz wave.

Step 2. We know that:

T =1

f(17.5)

Step 3.

T = 1f

= 110 Hz

= 0, 1 s

(17.6)

Step 4. The period of a 10 Hz wave is 0, 1 s.

1.129.6 Speed of a Transverse Wave

In Motion in One Dimension, we saw that speed was defined as

speed =distance traveled

time taken(17.7)

The distance between two successive peaks is 1 wavelength, λ. Thus in a time of 1 period, the wave will travel

1 wavelength in distance. Thus the speed of the wave, v, is:

v =distance traveled

time taken=

λ

T(17.8)

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However, f = 1T

. Therefore, we can also write:

v = λT

= λ · 1T

= λ · f(17.9)

We call this equation the wave equation. To summarise, we have that v = λ · f where

• v = speed in m · s−1

• λ = wavelength in m• f = frequency in Hz

Exercise 17.4: Speed of a Transverse Wave 1 When a particular

string is vibrated at a frequency of 10 Hz, a transverse wave of wave-

length 0, 25 m is produced. Determine the speed of the wave as it travels

along the string.

Solution to Exercise

Step 1. • frequency of wave: f = 10Hz• wavelength of wave: λ = 0, 25m

We are required to calculate the speed of the wave as it travels

along the string. All quantities are in SI units.

Step 2. We know that the speed of a wave is:

v = f · λ (17.10)

and we are given all the necessary quantities.

Step 3.

v = f · λ= (10 Hz) (0, 25m)

= 2, 5m · s−1

(17.11)

Step 4. The wave travels at 2, 5 m · s−1 along the string.

Exercise 17.5: Speed of a Transverse Wave 2 A cork on the surface of

a swimming pool bobs up and down once every second on some ripples.

The ripples have a wavelength of 20 cm. If the cork is 2 m from the edge

of the pool, how long does it take a ripple passing the cork to reach the

edge?

Solution to Exercise

Step 1. We are given:

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• frequency of wave: f = 1Hz• wavelength of wave: λ = 20 cm• distance of cork from edge of pool: D = 2m

We are required to determine the time it takes for a ripple to travel

between the cork and the edge of the pool.

The wavelength is not in SI units and should be converted.

Step 2. The time taken for the ripple to reach the edge of the pool is ob-

tained from:

t =D

v

(

from v =D

t

)

(17.12)

We know that

v = f · λ (17.13)

Therefore,

t =D

f · λ (17.14)

Step 3.

20 cm = 0, 2m (17.15)

Step 4.

t = Df ·λ

= 2 m(1 Hz)(0,2 m)

= 10 s

(17.16)

Step 5. A ripple passing the leaf will take 10 s to reach the edge of the pool.

The following video provides a summary of the concepts covered so far.

Khan academy video on waves - 1 (Video: P10041 )

Waves

1. When the particles of a medium move perpendicular to the direction of the wave motion, the wave is called

a ......... wave.

2. A transverse wave is moving downwards. In what direction do the particles in the medium move?

3. Consider the diagram below and answer the questions that follow:

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a. the wavelength of the wave is shown by letter .

b. the amplitude of the wave is shown by letter .

4. Draw 2 wavelengths of the following transverse waves on the same graph paper. Label all important values.

a. Wave 1: Amplitude = 1 cm, wavelength = 3 cm

b. Wave 2: Peak to trough distance (vertical) = 3 cm, peak to peak distance (horizontal) = 5 cm

5. You are given the transverse wave below.

Draw the following:

a. A wave with twice the amplitude of the given wave.

b. A wave with half the amplitude of the given wave.

c. A wave travelling at the same speed with twice the frequency of the given wave.

d. A wave travelling at the same speed with half the frequency of the given wave.

e. A wave with twice the wavelength of the given wave.

f. A wave with half the wavelength of the given wave.

g. A wave travelling at the same speed with twice the period of the given wave.

h. A wave travelling at the same speed with half the period of the given wave.

6. A transverse wave travelling at the same speed with an amplitude of 5 cm has a frequency of 15 Hz. The

horizontal distance from a crest to the nearest trough is measured to be 2,5 cm. Find the

a. period of the wave.

b. speed of the wave.

7. A fly flaps its wings back and forth 200 times each second. Calculate the period of a wing flap.

8. As the period of a wave increases, the frequency increases/decreases/does not change.

9. Calculate the frequency of rotation of the second hand on a clock.

10. Microwave ovens produce radiation with a frequency of 2 450 MHz (1 MHz = 106 Hz) and a wavelength of

0,122 m. What is the wave speed of the radiation?

11. Study the following diagram and answer the questions:

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a. Identify two sets of points that are in phase.

b. Identify two sets of points that are out of phase.

c. Identify any two points that would indicate a wavelength.

12. Tom is fishing from a pier and notices that four wave crests pass by in 8 s and estimates the distance be-

tween two successive crests is 4 m. The timing starts with the first crest and ends with the fourth. Calculate

the speed of the wave.

Find the answers with the shortcodes:

(1.) liq (2.) li4 (3.) li2 (4.) lr8 (5.) lr9 (6.) lrX

(7.) lrl (8.) lr5 (9.) lrN (10.) lrR (11.) lrn (12.) lrQ

1.130 Graphs of Particle Motion (Not in CAPS - Included for Interest)

(section shortcode: P10042 )

In Transverse Pulses, we saw that when a pulse moves through a medium, there are two different motions: the

motion of the particles of the medium and the motion of the pulse. These two motions are at right angles to each

other when the pulse is transverse. Since a transverse wave is a series of transverse pulses, the particle in the

medium and the wave move in exactly the same way as for the pulse.

When a transverse wave moves horizontally through the medium, the particles in the medium only move up and

down. We can see this in the figure below, which shows the motion of a single particle as a transverse wave

moves through the medium.

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TIP: A particle in the medium only moves up and down when a transverse wave moves

horizontally through the medium.

As in Transverse Pulses (Chapter 16), we can draw a graph of the particles’ position as a function of time. For

the wave shown in the above figure, we can draw the graph shown below.

The graph of the particle’s velocity as a function of time is obtained by taking the gradient of the position vs. time

graph. The graph of velocity vs. time for the position vs. time graph above, is shown in the graph below.

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The graph of the particle’s acceleration as a function of time is obtained by taking the gradient of the velocity vs.

time graph. The graph of acceleration vs. time for the position vs. time graph shown above, is shown below.

As for motion in one dimension, these graphs can be used to describe the motion of the particle in the medium.

This is illustrated in the worked examples below.

Exercise 17.6: Graphs of particle motion 1 The following graph

shows the position of a particle of a wave as a function of time.

1. Draw the corresponding velocity vs. time graph for the particle.

2. Draw the corresponding acceleration vs. time graph for the particle.

Solution to Exercise

Step 1. The y vs. t graph is given. The vy vs. t and ay vs. t graphs are

required.

Step 2. To find the velocity of the particle we need to find the gradient of the

y vs. t graph at each time. At point A the gradient is a maximum

and positive. At point B the gradient is zero. At point C the gradient

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is a maximum, but negative. At point D the gradient is zero. At point

E the gradient is a maximum and positive again.

Step 3. To find the acceleration of the particle we need to find the gradient

of the vy vs. t graph at each time. At point A the gradient is zero.

At point B the gradient is negative and a maximum. At point C the

gradient is zero. At point D the gradient is positive and a maximum.

At point E the gradient is zero.

1.130.1 Mathematical Description of Waves

If you look carefully at the pictures of waves you will notice that they look very much like sine or cosine functions.

This is correct. Waves can be described by trigonometric functions that are functions of time or of position.

Depending on which case we are dealing with the function will be a function of t or x. For example, a function of

position would be:

y (x) = Asin(

360x

λ+ φ

)

(17.17)

where A is the amplitude, λ the wavelength and φ is a phase shift. The phase shift accounts for the fact that the

wave at x = 0 does not start at the equilibrium position. A function of time would be:

y (t) = Asin

(

360t

T+ φ

)

(17.18)

where T is the period of the wave. Descriptions of the wave incorporate the amplitude, wavelength, frequency

or period and a phase shift.

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1.130.2 Graphs of Particle Motion

1. The following velocity vs. time graph for a particle in a wave is given.

a. Draw the corresponding position vs. time graph for the particle.

b. Draw the corresponding acceleration vs. time graph for the particle.

Find the answers with the shortcodes:

(1.) lrU

1.131 Standing Waves and Boundary Conditions (Not in CAPS - In-

cluded for Interest)

(section shortcode: P10043 )

1.131.1 Reflection of a Transverse Wave from a Fixed End

We have seen that when a pulse meets a fixed endpoint, the pulse is reflected, but it is inverted. Since a

transverse wave is a series of pulses, a transverse wave meeting a fixed endpoint is also reflected and the

reflected wave is inverted. That means that the peaks and troughs are swapped around.

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Figure 17.25: Reflection of a transverse wave from a fixed end.

1.131.2 Reflection of a Transverse Wave from a Free End

If transverse waves are reflected from an end, which is free to move, the waves sent down the string are reflected

but do not suffer a phase shift as shown in Figure 17.26.

Figure 17.26: Reflection of a transverse wave from a free end.

1.131.3 Standing Waves

What happens when a reflected transverse wave meets an incident transverse wave? When two waves move in

opposite directions, through each other, interference takes place. If the two waves have the same frequency and

wavelength then standing waves are generated.

Standing waves are so-called because they appear to be standing still.

Investigation : Creating Standing Waves

Tie a rope to a fixed object such that the tied end does not move. Continuously move the free end up and down

to generate firstly transverse waves and later standing waves.

We can now look closely how standing waves are formed. Figure 17.27 shows a reflected wave meeting an

incident wave.

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Figure 17.27: A reflected wave (solid line) approaches the incident wave (dashed line).

When they touch, both waves have an amplitude of zero:

Figure 17.28: A reflected wave (solid line) meets the incident wave (dashed line).

If we wait for a short time the ends of the two waves move past each other and the waves overlap. To find the

resultant wave, we add the two together.

Figure 17.29: A reflected wave (solid line) overlaps slightly with the incident wave (dashed line).

In this picture, we show the two waves as dotted lines and the sum of the two in the overlap region is shown as a

solid line:

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The important thing to note in this case is that there are some points where the two waves always destructively

interfere to zero. If we let the two waves move a little further we get the picture below:

Again we have to add the two waves together in the overlap region to see what the sum of the waves looks like.

In this case the two waves have moved half a cycle past each other but because they are completely out of phase

they cancel out completely.

When the waves have moved past each other so that they are overlapping for a large region the situation looks

like a wave oscillating in place. The following sequence of diagrams show what the resulting wave will look like. To

make it clearer, the arrows at the top of the picture show peaks where maximum positive constructive interference

is taking place. The arrows at the bottom of the picture show places where maximum negative interference is

taking place.

As time goes by the peaks become smaller and the troughs become shallower but they do not move.

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For an instant the entire region will look completely flat.

The various points continue their motion in the same manner.

Eventually the picture looks like the complete reflection through the x-axis of what we started with:

Then all the points begin to move back. Each point on the line is oscillating up and down with a different amplitude.

If we look at the overall result, we get a standing wave.

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Figure 17.39: A standing wave

If we superimpose the two cases where the peaks were at a maximum and the case where the same waves were

at a minimum we can see the lines that the points oscillate between. We call this the envelope of the standing

wave as it contains all the oscillations of the individual points. To make the concept of the envelope clearer let us

draw arrows describing the motion of points along the line.

Every point in the medium containing a standing wave oscillates up and down and the amplitude of the oscillations

depends on the location of the point. It is convenient to draw the envelope for the oscillations to describe the

motion. We cannot draw the up and down arrows for every single point!

NOTE: Standing waves can be a problem in for example indoor concerts where the

dimensions of the concert venue coincide with particular wavelengths. Standing waves

can appear as ‘feedback’, which would occur if the standing wave was picked up by

the microphones on stage and amplified.

1.131.4 Nodes and Anti-nodes

A node is a point on a wave where no displacement takes place at any time. In a standing wave, a node is a place

where two waves cancel out completely as the two waves destructively interfere in the same place. A fixed end

of a rope is a node. An anti-node is a point on a wave where maximum displacement takes place. In a standing

wave, an anti-node is a place where the two waves constructively interfere. Anti-nodes occur midway between

nodes. A free end of a rope is an anti-node.

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Definition: Node

A node is a point on a standing wave where no displacement takes place at any time. A

fixed end of a rope is a node.

Definition: Anti-Node

An anti-node is a point on standing a wave where maximum displacement takes place. A

free end of a rope is an anti-node.

TIP: The distance between two anti-nodes is only 12λ because it is the distance from a

peak to a trough in one of the waves forming the standing wave. It is the same as the

distance between two adjacent nodes. This will be important when we work out the

allowed wavelengths in tubes later. We can take this further because half-way between

any two anti-nodes is a node. Then the distance from the node to the anti-node is half

the distance between two anti-nodes. This is half of half a wavelength which is one

quarter of a wavelength, 14λ.

1.131.5 Wavelengths of Standing Waves with Fixed and Free Ends

There are many applications which make use of the properties of waves and the use of fixed and free ends. Most

musical instruments rely on the basic picture that we have presented to create specific sounds, either through

standing pressure waves or standing vibratory waves in strings.

The key is to understand that a standing wave must be created in the medium that is oscillating. There are

restrictions as to what wavelengths can form standing waves in a medium.

For example, if we consider a rope that can move in a pipe such that it can have

• both ends free to move (Case 1)

• one end free and one end fixed (Case 2)

• both ends fixed (Case 3).

Each of these cases is slightly different because the free or fixed end determines whether a node or anti-node

will form when a standing wave is created in the rope. These are the main restrictions when we determine the

wavelengths of potential standing waves. These restrictions are known as boundary conditions and must be met.

In the diagram below you can see the three different cases. It is possible to create standing waves with different

frequencies and wavelengths as long as the end criteria are met.

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The longer the wavelength the less the number of anti-nodes in the standing waves. We cannot have a standing

wave with no anti-nodes because then there would be no oscillations. We use n to number the anti-nodes. If

all of the tubes have a length L and we know the end constraints we can find the wavelength, λ, for a specific

number of anti-nodes.

One Node

Let’s work out the longest wavelength we can have in each tube, i.e. the case for n = 1.

Case 1: In the first tube, both ends must be anti-nodes, so we must place one node in the middle of the tube. We

know the distance from one anti-node to another is 12λ and we also know this distance is L. So we can equate

the two and solve for the wavelength:

12λ = L

λ = 2L(17.19)

Case 2: In the second tube, one end must be a node and the other must be an anti-node. Since we are looking

at the case with one node, we are forced to have it at the end. We know the distance from one node to another

is 12λ but we only have half this distance contained in the tube. So :

12

(

12λ

)

= L

λ = 4L(17.20)

Case 3: Here both ends are closed and so we must have two nodes so it is impossible to construct a case with

only one node.

Two Nodes

Next we determine which wavelengths could be formed if we had two nodes. Remember that we are dividing the

tube up into smaller and smaller segments by having more nodes so we expect the wavelengths to get shorter.

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Case 1: Both ends are open and so they must be anti-nodes. We can have two nodes inside the tube only if

we have one anti-node contained inside the tube and one on each end. This means we have 3 anti-nodes in

the tube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength

between the left side and the middle and another half wavelength between the middle and the right side so there

must be one wavelength inside the tube. The safest thing to do is work out how many half wavelengths there are

and equate this to the length of the tube L and then solve for λ.

2(

12λ

)

= L

λ = L(17.21)

Case 2: We want to have two nodes inside the tube. The left end must be a node and the right end must be an

anti-node. We can have one node inside the tube as drawn above. Again we can count the number of distances

between adjacent nodes or anti-nodes. If we start from the left end we have one half wavelength between the end

and the node inside the tube. The distance from the node inside the tube to the right end which is an anti-node

is half of the distance to another node. So it is half of half a wavelength. Together these add up to the length of

the tube:

12λ+ 1

2

(

12λ

)

= L

24λ+ 1

4λ = L

34λ = L

λ = 43L

(17.22)

Case 3: In this case both ends have to be nodes. This means that the length of the tube is one half wavelength:

So we can equate the two and solve for the wavelength:

12λ = L

λ = 2L(17.23)

TIP: If you ever calculate a longer wavelength for more nodes you have made a mis-

take. Remember to check if your answers make sense!

Three Nodes

To see the complete pattern for all cases we need to check what the next step for case 3 is when we have an

additional node. Below is the diagram for the case where n = 3.

Case 1: Both ends are open and so they must be anti-nodes. We can have three nodes inside the tube only if

we have two anti-nodes contained inside the tube and one on each end. This means we have 4 anti-nodes in

the tube. The distance between any two anti-nodes is half a wavelength. This means there is half wavelength

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1.131. STANDING WAVES AND BOUNDARY CONDITIONS (NOT IN CAPS - INCLUDED FOR INTEREST)CHAPTER 1. TRANSVERSE WAVES

between every adjacent pair of anti-nodes. We count how many gaps there are between adjacent anti-nodes to

determine how many half wavelengths there are and equate this to the length of the tube L and then solve for λ.

3(

12λ

)

= L

λ = 23L

(17.24)

Case 2: We want to have three nodes inside the tube. The left end must be a node and the right end must

be an anti-node, so there will be two nodes between the ends of the tube. Again we can count the number of

distances between adjacent nodes or anti-nodes, together these add up to the length of the tube. Remember

that the distance between the node and an adjacent anti-node is only half the distance between adjacent nodes.

So starting from the left end we count 3 nodes, so 2 half wavelength intervals and then only a node to anti-node

distance:

2(

12λ

)

+ 12

(

12λ

)

= L

λ+ 14λ = L

54λ = L

λ = 45L

(17.25)

Case 3: In this case both ends have to be nodes. With one node in between there are two sets of adjacent

nodes. This means that the length of the tube consists of two half wavelength sections:

2(

12λ

)

= L

λ = L(17.26)

1.131.6 Superposition and Interference

If two waves meet interesting things can happen. Waves are basically collective motion of particles. So when two

waves meet they both try to impose their collective motion on the particles. This can have quite different results.

If two identical (same wavelength, amplitude and frequency) waves are both trying to form a peak then they are

able to achieve the sum of their efforts. The resulting motion will be a peak which has a height which is the sum of

the heights of the two waves. If two waves are both trying to form a trough in the same place then a deeper trough

is formed, the depth of which is the sum of the depths of the two waves. Now in this case, the two waves have

been trying to do the same thing, and so add together constructively. This is called constructive interference.

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If one wave is trying to form a peak and the other is trying to form a trough, then they are competing to do different

things. In this case, they can cancel out. The amplitude of the resulting wave will depend on the amplitudes of

the two waves that are interfering. If the depth of the trough is the same as the height of the peak nothing will

happen. If the height of the peak is bigger than the depth of the trough, a smaller peak will appear. And if the

trough is deeper then a less deep trough will appear. This is destructive interference.

Superposition and Interference

1. For each labelled point, indicate whether constructive or destructive interference takes place at that point.

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1.131. STANDING WAVES AND BOUNDARY CONDITIONS (NOT IN CAPS - INCLUDED FOR INTEREST)CHAPTER 1. TRANSVERSE WAVES

2. A ride at the local amusement park is called "Standing on Standing Waves". Which position (a node or an

antinode) on the ride would give the greatest thrill?

3. How many nodes and how many anti-nodes appear in the standing wave below?

4. For a standing wave on a string, you are given three statements:

A. you can have any λ and any f as long as the relationship, v = λ · f is satisfied.

B. only certain wavelengths and frequencies are allowed

C. the wave velocity is only dependent on the medium

Which of the statements are true:

a. A and C only

b. B and C only

c. A, B, and C

d. none of the above

5. Consider the diagram below of a standing wave on a string 9 m long that is tied at both ends. The wave

velocity in the string is 16 m · s−1. What is the wavelength?

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CHAPTER 1. TRANSVERSE WAVES 1.132. SUMMARY

Find the answers with the shortcodes:

(1.) lrP (2.) lrE (3.) lrm (4.) lry (5.) lrG

1.132 Summary

(section shortcode: P10044 )

1. A wave is formed when a continuous number of pulses are transmitted through a medium.

2. A peak is the highest point a particle in the medium rises to.

3. A trough is the lowest point a particle in the medium sinks to.

4. In a transverse wave, the particles move perpendicular to the motion of the wave.

5. The amplitude is the maximum distance from equilibrium position to a peak (or trough), or the maximum

displacement of a particle in a wave from its position of rest.

6. The wavelength (λ) is the distance between any two adjacent points on a wave that are in phase. It is

measured in metres.

7. The period (T ) of a wave is the time it takes a wavelength to pass a fixed point. It is measured in seconds

(s).

8. The frequency (f ) of a wave is how many waves pass a point in a second. It is measured in hertz (Hz) or

s−1.

9. Frequency: f = 1T

10. Period: T = 1f

11. Speed: v = fλ or v = λT

.

12. When a wave is reflected from a fixed end, the resulting wave will move back through the medium, but will

be inverted. When a wave is reflected from a free end, the waves are reflected, but not inverted.

1.133 Exercises

(section shortcode: P10045 )

1. A standing wave is formed when:

a. a wave refracts due to changes in the properties of the medium

b. a wave reflects off a canyon wall and is heard shortly after it is formed

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1.133. EXERCISES CHAPTER 1. TRANSVERSE WAVES

c. a wave refracts and reflects due to changes in the medium

d. two identical waves moving different directions along the same medium interfere

2. How many nodes and anti-nodes are shown in the diagram?

3. Draw a transverse wave that is reflected from a fixed end.

4. Draw a transverse wave that is reflected from a free end.

5. A wave travels along a string at a speed of 1, 5m · s−1. If the frequency of the source of the wave is 7,5 Hz,

calculate:

a. the wavelength of the wave

b. the period of the wave

6. Water waves crash against a seawall around the harbour. Eight waves hit the seawall in 5 s. The distance

between successive troughs is 9 m. The height of the waveform trough to crest is 1,5 m.

a. How many complete waves are indicated in the sketch?

b. Write down the letters that indicate any TWO points that are:

i. in phase

ii. out of phase

iii. Represent ONE wavelength.

c. Calculate the amplitude of the wave.

d. Show that the period of the wave is 0,67 s.

e. Calculate the frequency of the waves.

f. Calculate the velocity of the waves.

Find the answers with the shortcodes:

(1.) lrV (2.) lrp (3.) lrd (4.) lrv (5.) lrw (6.) lrf

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Longitudinal Waves

1.134 Introduction

(section shortcode: P10046 )

We have already studied transverse pulses and waves. In this chapter we look at another type of wave called

longitudinal waves. In transverse waves, the motion of the particles in the medium were perpendicular to the

direction of the wave. In longitudinal waves, the particles in the medium move parallel (in the same direction

as) to the motion of the wave. Examples of transverse waves are water waves or light waves. An example of a

longitudinal wave is a sound wave.

1.135 What is a longitudinal wave?

(section shortcode: P10047 )

Definition: Longitudinal waves

A longitudinal wave is a wave where the particles in the medium move parallel to the

direction of propagation of the wave.

When we studied transverse waves we looked at two different motions: the motion of the particles of the medium

and the motion of the wave itself. We will do the same for longitudinal waves.

The question is how do we construct such a wave?

To create a transverse wave, we flick the end of for example a rope up and down. The particles move up and

down and return to their equilibrium position. The wave moves from left to right and will be displaced.

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1.135. WHAT IS A LONGITUDINAL WAVE? CHAPTER 1. LONGITUDINAL WAVES

A longitudinal wave is seen best in a spring that is hung from a ceiling. Do the following investigation to find out

more about longitudinal waves.

1.135.1 Investigation : Investigating longitudinal waves

1. Take a spring and hang it from the ceiling. Pull the free end of the spring and release it. Observe what

happens.

2. In which direction does the disturbance move?

3. What happens when the disturbance reaches the ceiling?

4. Tie a ribbon to the middle of the spring. Watch carefully what happens to the ribbon when the free end of

the spring is pulled and released. Describe the motion of the ribbon.

From the investigation you will have noticed that the disturbance moves parallel to the direction in which the spring

was pulled. The spring was pulled down and the wave moved up and down. The ribbon in the investigation

represents one particle in the medium. The particles in the medium move in the same direction as the wave.

The ribbon moves from rest upwards, then back to its original position, then down and then back to its original

position.

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CHAPTER 1. LONGITUDINAL WAVES 1.136. CHARACTERISTICS OF LONGITUDINAL WAVES

Figure 18.3: Longitudinal wave through a spring

1.136 Characteristics of Longitudinal Waves

(section shortcode: P10048 )

As in the case of transverse waves the following properties can be defined for longitudinal waves: wavelength,

amplitude, period, frequency and wave speed. However instead of peaks and troughs, longitudinal waves have

compressions and rarefactions.

Definition: Compression

A compression is a region in a longitudinal wave where the particles are closest together.

Definition: Rarefaction

A rarefaction is a region in a longitudinal wave where the particles are furthest apart.

1.136.1 Compression and Rarefaction

As seen in Figure 18.4, there are regions where the medium is compressed and other regions where the medium

is spread out in a longitudinal wave.

The region where the medium is compressed is known as a compression and the region where the medium is

spread out is known as a rarefaction.

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1.136. CHARACTERISTICS OF LONGITUDINAL WAVES CHAPTER 1. LONGITUDINAL WAVES

Figure 18.4: Compressions and rarefactions on a longitudinal wave

1.136.2 Wavelength and Amplitude

Definition: Wavelength

The wavelength in a longitudinal wave is the distance between two consecutive points

that are in phase.

The wavelength in a longitudinal wave refers to the distance between two consecutive compressions or between

two consecutive rarefactions.

Definition: Amplitude

The amplitude is the maximum displacement from equilibrium. For a longitudinal wave

which is a pressure wave this would be the maximum increase (or decrease) in pressure

from the equilibrium pressure that is cause when a peak (or trough) passes a point.

Figure 18.5: Wavelength on a longitudinal wave

The amplitude is the distance from the equilibrium position of the medium to a compression or a rarefaction.

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CHAPTER 1. LONGITUDINAL WAVES 1.136. CHARACTERISTICS OF LONGITUDINAL WAVES

1.136.3 Period and Frequency

Definition: Period

The period of a wave is the time taken by the wave to move one wavelength.

Definition: Frequency

The frequency of a wave is the number of wavelengths per second.

The period of a longitudinal wave is the time taken by the wave to move one wavelength. As for transverse waves,

the symbol T is used to represent period and period is measured in seconds (s).

The frequencyf of a wave is the number of wavelengths per second. Using this definition and the fact that the

period is the time taken for 1 wavelength, we can define:

f =1

T(18.1)

or alternately,

T =1

f(18.2)

1.136.4 Speed of a Longitudinal Wave

The speed of a longitudinal wave is defined as:

v = f · λ (18.3)

where

• v = speed in m · s−1

• f = frequency in Hz• λ = wavelength in m

Exercise 18.1: Speed of longitudinal waves The musical note “A” is

a sound wave. The note has a frequency of 440 Hz and a wavelength of

0,784 m. Calculate the speed of the musical note.

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1.136. CHARACTERISTICS OF LONGITUDINAL WAVES CHAPTER 1. LONGITUDINAL WAVES

Solution to Exercise

Step 1.

f = 440 Hz

λ = 0, 784 m(18.4)

We need to calculate the speed of the musical note “A”.

Step 2. We are given the frequency and wavelength of the note. We can

therefore use:

v = f · λ (18.5)

Step 3.

v = f · λ= (440 Hz) (0, 784m)

= 345m · s−1

(18.6)

Step 4. The musical note “A” travels at 345 m · s−1.

Exercise 18.2: Speed of longitudinal waves A longitudinal wave trav-

els into a medium in which its speed increases. How does this affect its...

(write only increases, decreases, stays the same).

1. period?

2. wavelength?

Solution to Exercise

Step 1. We need to determine how the period and wavelength of a longitu-

dinal wave change when its speed increases.

Step 2. We need to find the link between period, wavelength and wave

speed.

Step 3. We know that the frequency of a longitudinal wave is dependent

on the frequency of the vibrations that lead to the creation of the

longitudinal wave. Therefore, the frequency is always unchanged,

irrespective of any changes in speed. Since the period is the in-

verse of the frequency, the period remains the same.

Step 4. The frequency remains unchanged. According to the wave equa-

tion

v = fλ (18.7)

if f remains the same and v increases, then λ, the wavelength,

must also increase.

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CHAPTER 1. LONGITUDINAL WAVES 1.137. SOUND WAVES

1.137 Sound Waves

(section shortcode: P10049 )

Sound waves coming from a tuning fork are caused by the vibrations of the tuning fork which push against the

air particles in front of it. As the air particles are pushed together a compression is formed. The particles behind

the compression move further apart causing a rarefaction. As the particles continue to push against each other,

the sound wave travels through the air. Due to this motion of the particles, there is a constant variation in the

pressure in the air. Sound waves are therefore pressure waves. This means that in media where the particles

are closer together, sound waves will travel quicker.

Sound waves travel faster through liquids, like water, than through the air because water is denser than air (the

particles are closer together). Sound waves travel faster in solids than in liquids.

Figure 18.6: Sound waves are pressure waves and need a medium through which to travel.

TIP: A sound wave is different from a light wave.

• A sound wave is produced by an oscillating object while a light wave is not.

Also, because a sound wave is a mechanical wave (i.e. that it needs a medium) it is

not capable of traveling through a vacuum, whereas a light wave can travel through a

vacuum.

TIP: A sound wave is a pressure wave. This means that regions of high pressure (com-

pressions) and low pressure (rarefactions) are created as the sound source vibrates.

These compressions and rarefactions arise because the source vibrates longitudinally

and the longitudinal motion of air produces pressure fluctuations.

Sound will be studied in more detail in Sound (Chapter 19).

1.138 Seismic Waves - (Not Included in CAPS - Included for Interest)

(section shortcode: P10050 )

Seismic waves are waves from vibrations in the Earth (core, mantle, oceans). Seismic waves also occur on other

planets, for example the moon and can be natural (due to earthquakes, volcanic eruptions or meteor strikes)

or man-made (due to explosions or anything that hits the earth hard). Seismic P-waves (P for pressure) are

longitudinal waves which can travel through solid and liquid.

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1.139. GRAPHS OF PARTICLE POSITION, DISPLACEMENT, VELOCITY AND ACCELERATION (NOT

INCLUDED IN CAPS - INCLUDED FOR COMPLETENESS) CHAPTER 1. LONGITUDINAL WAVES

1.139 Graphs of Particle Position, Displacement, Velocity and Acceler-

ation (Not Included in CAPS - Included for Completeness)

(section shortcode: P10051 )

When a longitudinal wave moves through the medium, the particles in the medium only move back and forth

relative to the direction of motion of the wave. We can see this in Figure 18.7 which shows the motion of the

particles in a medium as a longitudinal wave moves through the medium.

Figure 18.7: Positions of particles in a medium at different times as a longitudinal wave moves through it. The

wave moves to the right. The dashed line shows the equilibrium position of particle 0.

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CHAPTER 1. LONGITUDINAL WAVES

1.139. GRAPHS OF PARTICLE POSITION, DISPLACEMENT, VELOCITY AND ACCELERATION (NOT

INCLUDED IN CAPS - INCLUDED FOR COMPLETENESS)

TIP: A particle in the medium only moves back and forth when a longitudinal wave

moves through the medium.

We can draw a graph of the particle’s change in position from its starting point as a function of time. For the wave

shown in Figure 18.7, we can draw the graph shown in Figure 18.8 for particle 0. The graph for each of the other

particles will be identical.

Figure 18.8: Graph of particle displacement as a function of time for the longitudinal wave shown in Figure 18.7.

The graph of the particle’s velocity as a function of time is obtained by taking the gradient of the position vs. time

graph. The graph of velocity vs. time for the position vs. time graph shown in Figure 18.8 is shown is Figure 18.9.

Figure 18.9: Graph of velocity as a function of time.

The graph of the particle’s acceleration as a function of time is obtained by taking the gradient of the velocity vs.

time graph. The graph of acceleration vs. time for the position vs. time graph shown in Figure 18.8 is shown is

Figure 18.10.

Figure 18.10: Graph of acceleration as a function of time.

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1.140. SUMMARY - LONGITUDINAL WAVES CHAPTER 1. LONGITUDINAL WAVES

1.140 Summary - Longitudinal Waves

(section shortcode: P10052 )

1. A longitudinal wave is a wave where the particles in the medium move parallel to the direction in which the

wave is travelling.

2. Longitudinal waves consist of areas of higher pressure, where the particles in the medium are closest

together (compressions) and areas of lower pressure, where the particles in the medium are furthest apart

(rarefactions).

3. The wavelength of a longitudinal wave is the distance between two consecutive compressions, or two con-

secutive rarefactions.

4. The relationship between the period (T ) and frequency (f ) is given by

T =1

for f =

1

T(18.8)

5. The relationship between wave speed (v), frequency (f ) and wavelength (λ) is given by

v = fλ (18.9)

6. Graphs of position vs time, velocity vs time and acceleration vs time can be drawn and are summarised in

figures

7. Sound waves are examples of longitudinal waves. The speed of sound depends on the medium, temper-

ature and pressure. Sound waves travel faster in solids than in liquids, and faster in liquids than in gases.

Sound waves also travel faster at higher temperatures and higher pressures.

1.141 Exercises - Longitudinal Waves

(section shortcode: P10053 )

1. Which of the following is not a longitudinal wave?

a. seismic P-wave

b. light

c. sound

d. ultrasound

2. Which of the following media can sound not travel through?

a. solid

b. liquid

c. gas

d. vacuum

3. Select a word from Column B that best fits the description in Column A:

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CHAPTER 1. LONGITUDINAL WAVES 1.141. EXERCISES - LONGITUDINAL WAVES

Column A Column B

waves in the air caused by vibrations longitudinal waves

waves that move in one direction, but medium moves in another frequency

waves and medium that move in the same direction white noise

the distance between consecutive points of a wave which are in phase amplitude

how often a single wavelength goes by sound waves

half the difference between high points and low points of waves standing waves

the distance a wave covers per time interval transverse waves

the time taken for one wavelength to pass a point wavelength

music

sounds

wave speed

Table 18.1

4. A longitudinal wave has a crest to crest distance of 10 m. It takes the wave 5 s to pass a point.

a. What is the wavelength of the longitudinal wave?

b. What is the speed of the wave?

5. A flute produces a musical sound travelling at a speed of 320 m · s−1. The frequency of the note is 256 Hz.

Calculate:

a. the period of the note

b. the wavelength of the note

6. A person shouts at a cliff and hears an echo from the cliff 1 s later. If the speed of sound is 344 m · s−1, how

far away is the cliff?

7. A wave travels from one medium to another and the speed of the wave decreases. What will the effect be

on the ... (write only increases, decreases or remains the same)

a. wavelength?

b. period?

Find the answers with the shortcodes:

(1.) l2d (2.) l2v (3.) l2w (4.) l2f (5.) l2G (6.) l27

(7.) l2A

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1.141. EXERCISES - LONGITUDINAL WAVES CHAPTER 1. LONGITUDINAL WAVES

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Sound

1.142 Introduction

(section shortcode: P10054 )

Now that we have studied the basics of longitudinal waves, we are ready to study sound waves in detail.

Have you ever thought about how amazing your sense of hearing is? It is actually pretty remarkable. There are

many types of sounds: a car horn, a laughing baby, a barking dog, and somehow your brain can sort it all out.

Though it seems complicated, it is rather simple to understand once you learn a very simple fact. Sound is a

wave. So you can use everything you know about waves to explain sound.

1.143 Characteristics of a Sound Wave

(section shortcode: P10055 )

Since sound is a wave, we can relate the properties of sound to the properties of a wave. The basic properties of

sound are: pitch, loudness and tone.

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1.143. CHARACTERISTICS OF A SOUND WAVE CHAPTER 1. SOUND

(a)

(b)

(c)

Figure 19.1: Pitch and loudness of sound. Sound B has a lower pitch (lower frequency) than Sound A and is

softer (smaller amplitude) than Sound C.

1.143.1 Pitch

The frequency of a sound wave is what your ear understands as pitch. A higher frequency sound has a higher

pitch, and a lower frequency sound has a lower pitch. In Figure 19.1 sound A has a higher pitch than sound B.

For instance, the chirp of a bird would have a high pitch, but the roar of a lion would have a low pitch.

The human ear can detect a wide range of frequencies. Frequencies from 20 to 20 000 Hz are audible to the

human ear. Any sound with a frequency below 20 Hz is known as an infrasound and any sound with a frequency

above 20 000 Hz is known as an ultrasound.

Table 19.1 lists the hearing ranges of some common animals compared to humans.

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CHAPTER 1. SOUND 1.144. SPEED OF SOUND

lower frequency (Hz) upper frequency (Hz)

Humans 20 20 000

Dogs 50 45 000

Cats 45 85 000

Bats 20 120 000

Dolphins 0,25 200 000

Elephants 5 10 000

Table 19.1: Range of frequencies

Investigation : Range of Wavelengths

Using the information given in Table 19.1, calculate the lower and upper wavelengths that each species can hear.

Assume the speed of sound in air is 344 m · s−1.

1.143.2 Loudness

The amplitude of a sound wave determines its loudness or volume. A larger amplitude means a louder sound,

and a smaller amplitude means a softer sound. In Figure 19.1 sound C is louder than sound B. The vibration of

a source sets the amplitude of a wave. It transmits energy into the medium through its vibration. More energetic

vibration corresponds to larger amplitude. The molecules move back and forth more vigorously.

The loudness of a sound is also determined by the sensitivity of the ear. The human ear is more sensitive to

some frequencies than to others. The volume we receive thus depends on both the amplitude of a sound wave

and whether its frequency lies in a region where the ear is more or less sensitive.

1.143.3 Tone

Tone is a measure of the quality of the sound wave. For example, the quality of the sound produced in a particular

musical instruments depends on which harmonics are superposed and in which proportions. The harmonics are

determined by the standing waves that are produced in the instrument. For general interest see Physics of music,

which explains the physics of music in greater detail.

The quality (timbre) of the sound heard depends on the pattern of the incoming vibrations, i.e. the shape of the

sound wave. The more irregular the vibrations, the more jagged is the shape of the sound wave and the harsher

is the sound heard.

1.144 Speed of Sound

(section shortcode: P10056 )

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1.145. PHYSICS OF THE EAR AND HEARING CHAPTER 1. SOUND

The speed of sound depends on the medium the sound is travelling in. Sound travels faster in solids than in

liquids, and faster in liquids than in gases. This is because the density of solids is higher than that of liquids which

means that the particles are closer together. Sound can be transmitted more easily.

The speed of sound also depends on the temperature of the medium. The hotter the medium is, the faster its

particles move and therefore the quicker the sound will travel through the medium. When we heat a substance,

the particles in that substance have more kinetic energy and vibrate or move faster. Sound can therefore be

transmitted more easily and quickly in hotter substances.

Sound waves are pressure waves. The speed of sound will therefore be influenced by the pressure of the medium

through which it is travelling. At sea level the air pressure is higher than high up on a mountain. Sound will travel

faster at sea level where the air pressure is higher than it would at places high above sea level.

Definition: Speed of sound

The speed of sound in air, at sea level, at a temperature of 21C and under normal atmo-

spheric conditions, is 344 m · s−1.

1.144.1 Sound frequency and amplitude

Study the following diagram representing a musical note. Redraw the diagram for a note

1. with a higher pitch

2. that is louder

3. that is softer

Find the answers with the shortcodes:

(1.) l2o

1.145 Physics of the Ear and Hearing

(section shortcode: P10057 )

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CHAPTER 1. SOUND 1.146. ULTRASOUND

Figure 19.3: Diagram of the human ear.

The human ear is divided into three main sections: the outer, middle, and inner ear. Let’s follow the journey of

a sound wave from the pinna (outermost part) to the auditory nerve (innermost part) which transmits a signal

to the brain. The pinna is the part of the ear we typically think of when we refer to the ear. Its main function

is to collect and focus an incident sound wave. The wave then travels through the ear canal until it meets the

eardrum. The pressure fluctuations of the sound wave make the eardrum vibrate. The three very small bones of

the middle ear, the malleus (hammer), the incus (anvil), and the stapes (stirrup), transmit the signal through to

the elliptical window. The elliptical window is the beginning of the inner ear. From the elliptical window the sound

waves are transmitted through the liquid in the inner ear and interpreted as sounds by the brain. The inner ear,

made of the semicircular canals, the cochlea, and the auditory nerve, is filled with fluid. The fluid allows the body

to detect quick movements and maintain balance. The snail-shaped cochlea is covered in nerve cells. There

are more than 25 000 hairlike nerve cells. Different nerve cells vibrate with different frequencies. When a nerve

cell vibrates, it releases electrical impulses to the auditory nerve. The impulses are sent to the brain through the

auditory nerve and understood as sound.

1.146 Ultrasound

(section shortcode: P10058 )

Ultrasound is sound with a frequency that is higher than 20 kHz. Some animals, such as dogs, dolphins, and

bats, have an upper limit that is greater than that of the human ear and can hear ultrasound.

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1.147. SONAR CHAPTER 1. SOUND

Application Lowest Frequency (kHz) Highest Frequency (kHz)

Cleaning (e.g. Jewelery) 20 40

Material testing for flaws 50 500

Welding of plastics 15 40

Tumour ablation 250 2000

Table 19.2: Different uses of ultrasound and the frequencies applicable.

The most common use of ultrasound is to create images, and has industrial and medical applications. The use

of ultrasound to create images is based on the reflection and transmission of a wave at a boundary. When an

ultrasound wave travels inside an object that is made up of different materials such as the human body, each time

it encounters a boundary, e.g. between bone and muscle, or muscle and fat, part of the wave is reflected and

part of it is transmitted. The reflected rays are detected and used to construct an image of the object.

Ultrasound in medicine can visualise muscle and soft tissue, making them useful for scanning the organs, and is

commonly used during pregnancy. Ultrasound is a safe, non-invasive method of looking inside the human body.

Ultrasound sources may be used to generate local heating in biological tissue, with applications in physical

therapy and cancer treatment. Focussed ultrasound sources may be used to break up kidney stones.

Ultrasonic cleaners, sometimes called supersonic cleaners, are used at frequencies from 20-40 kHz for jewellery,

lenses and other optical parts, watches, dental instruments, surgical instruments and industrial parts. These

cleaners consist of containers with a fluid in which the object to be cleaned is placed. Ultrasonic waves are

then sent into the fluid. The main mechanism for cleaning action in an ultrasonic cleaner is actually the energy

released from the collapse of millions of microscopic bubbles occurring in the liquid of the cleaner.

NOTE: Ultrasound generator/speaker systems are sold with claims that they frighten

away rodents and insects, but there is no scientific evidence that the devices work;

controlled tests have shown that rodents quickly learn that the speakers are harmless.

In echo-sounding the reflections from ultrasound pulses that are bounced off objects (for example the bottom of

the sea, fish etc.) are picked up. The reflections are timed and since their speed is known, the distance to the

object can be found. This information can be built into a picture of the object that reflects the ultrasound pulses.

1.147 SONAR

(section shortcode: P10059 )

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CHAPTER 1. SOUND 1.147. SONAR

Ships on the ocean make use of the reflecting properties of sound waves to determine the depth of the ocean.

A sound wave is transmitted and bounces off the seabed. Because the speed of sound is known and the time

lapse between sending and receiving the sound can be measured, the distance from the ship to the bottom of

the ocean can be determined, This is called sonar, which stands from Sound Navigation And Ranging.

1.147.1 Echolocation

Animals like dolphins and bats make use of sounds waves to find their way. Just like ships on the ocean, bats

use sonar to navigate. Ultrasound waves that are sent out are reflected off the objects around the animal. Bats,

or dolphins, then use the reflected sounds to form a “picture” of their surroundings. This is called echolocation.

Exercise 19.1: SONAR A ship sends a signal to the bottom of the

ocean to determine the depth of the ocean. The speed of sound in sea

water is 1450 m.s−1. If the signal is received 1,5 seconds later, how deep

is the ocean at that point?

Solution to Exercise

Step 1.

s = 1450 m.s−1

t = 1, 5 s there and back

∴ t = 0, 75 s one way

D = ?

(19.1)

Step 2.

Distance = speed× time

D = s× t

= 1450m.s−1 × 0, 75s

= 1087, 5 m

(19.2)

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1.148. INTENSITY OF SOUND (NOT INCLUDED IN CAPS - ADVANCED) CHAPTER 1. SOUND

1.148 Intensity of Sound (Not Included in CAPS - Advanced)

(section shortcode: P10060 )

IMPORTANT: This section is more advanced than required and is best revisited for

interest only when you are comfortable with concepts like power and logarithms.

Intensity is one indicator of amplitude. Intensity is the energy transmitted over a unit of area each second.

1.148.1 Intensity

Intensity is defined as:

Intensity =energy

time× area=

power

area(19.3)

By the definition of intensity, we can see that the units of intensity are

Joules

s ·m2=

Watts

m2(19.4)

The unit of intensity is the decibel (symbol: dB). This reduces to an SI equivalent of W ·m−2.

The average threshold of hearing is 10−12 W ·m−2. Below this intensity, the sound is too soft for the ear to hear.

The threshold of pain is 1.0 W · m−2. Above this intensity a sound is so loud it becomes uncomfortable for the

ear.

Notice that there is a factor of 1012 between the thresholds of hearing and pain. This is one reason we define the

decibel (dB) scale.

1.148.2 dB Scale

The intensity in dB of a sound of intensity I, is given by:

β = 10 logI

IoIo = 10−12 W ·m−2 (19.5)

In this way we can compress the whole hearing intensity scale into a range from 0 dB to 120 dB.

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CHAPTER 1. SOUND 1.149. SUMMARY

Source Intensity (dB) Times greater than hearing threshold

Rocket Launch 180 1018

Jet Plane 140 1014

Threshold of Pain 120 1012

Rock Band 110 1011

Subway Train 90 109

Factory 80 108

City Traffic 70 107

Normal Conversation 60 106

Library 40 104

Whisper 20 102

Threshold of hearing 0 0

Table 19.3: Examples of sound intensities.

Notice that there are sounds which exceed the threshold of pain. Exposure to these sounds can cause immediate

damage to hearing. In fact, exposure to sounds from 80 dB and above can damage hearing over time. Measures

can be taken to avoid damage, such as wearing earplugs or ear muffs. Limiting exposure time and increasing

distance between you and the source are also important steps for protecting your hearing.

1.148.3 Discussion : Importance of Safety Equipment

Working in groups of 5, discuss the importance of safety equipment such as ear protectors for workers in loud

environments, e.g. those who use jack hammers or direct aeroplanes to their parking bays. Write up your con-

clusions in a one page report. Some prior research into the importance of safety equipment might be necessary

to complete this group discussion.

1.149 Summary

(section shortcode: P10061 )

1. Sound waves are longitudinal waves

2. The frequency of a sound is an indication of how high or low the pitch of the sound is.

3. The human ear can hear frequencies from 20 to 20 000 Hz. Infrasound waves have frequencies lower than

20 Hz. Ultrasound waves have frequencies higher than 20 000 Hz.

4. The amplitude of a sound determines its loudness or volume.

5. The tone is a measure of the quality of a sound wave.

6. The speed of sound in air is around 340 m · s−1. It is dependent on the temperature, height above sea level

and the phase of the medium through which it is travelling.

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1.150. EXERCISES CHAPTER 1. SOUND

7. Sound travels faster when the medium is hot.

8. Sound travels faster in a solid than a liquid and faster in a liquid than in a gas.

9. Sound travels faster at sea level where the air pressure is higher.

10. The intensity of a sound is the energy transmitted over a certain area. Intensity is a measure of frequency.

11. Ultrasound can be used to form pictures of things we cannot see, like unborn babies or tumors.

12. Echolocation is used by animals such as dolphins and bats to “see” their surroundings by using ultrasound.

13. Ships use sonar to determine how deep the ocean is or to locate shoals of fish.

1.150 Exercises

(section shortcode: P10062 )

1. Choose a word from column B that best describes the concept in column A.

Column A Column B

pitch of sound amplitude

loudness of sound frequency

quality of sound speed

waveform

Table 19.4

2. A tuning fork, a violin string and a loudspeaker are producing sounds. This is because they are all in a state

of:

a. compression

b. rarefaction

c. rotation

d. tension

e. vibration

3. What would a drummer do to make the sound of a drum give a note of lower pitch?

a. hit the drum harder

b. hit the drum less hard

c. hit the drum near the edge

d. loosen the drum skin

e. tighten the drum skin

4. What is the approximate range of audible frequencies for a healthy human?

a. 0.2 Hz → 200 Hz

b. 2 Hz → 2 000 Hz

c. 20 Hz → 20 000 Hz

d. 200 Hz → 200 000 Hz

e. 2 000 Hz → 2 000 000 Hz

5. X and Y are different wave motions. In air, X travels much faster than Y but has a much shorter wavelength.

Which types of wave motion could X and Y be?

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CHAPTER 1. SOUND 1.150. EXERCISES

X Y

A microwaves red light

B radio infra red

C red light sound

D sound ultraviolet

E ultraviolet radio

Table 19.5

6. Astronauts are in a spaceship orbiting the moon. They see an explosion on the surface of the moon. Why

can they not hear the explosion?

a. explosions do not occur in space

b. sound cannot travel through a vacuum

c. sound is reflected away from the spaceship

d. sound travels too quickly in space to affect the ear drum

e. the spaceship would be moving at a supersonic speed

7. A man stands between two cliffs as shown in the diagram and claps his hands once.

Assuming that the velocity of sound is 330 m · s−1, what will be the time interval between the two loudest

echoes?

a. 23 s

b. 16 s

c. 56 s

d. 1 s

e. 13 s

8. A dolphin emits an ultrasonic wave with frequency of 0,15 MHz. The speed of the ultrasonic wave in water

is 1500 m · s−1. What is the wavelength of this wave in water?

a. 0,1 mm

b. 1 cm

c. 10 cm

d. 10 m

e. 100 m

9. The amplitude and frequency of a sound wave are both increased. How are the loudness and pitch of the

sound affected?

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1.150. EXERCISES CHAPTER 1. SOUND

loudness pitch

A increased raised

B increased unchanged

C increased lowered

D decreased raised

E decreased lowered

Table 19.6

10. A jet fighter travels slower than the speed of sound. Its speed is said to be:

a. Mach 1

b. supersonic

c. subsonic

d. hypersonic

e. infrasonic

11. A sound wave is different from a light wave in that a sound wave is:

a. produced by a vibrating object and a light wave is not.

b. not capable of traveling through a vacuum.

c. not capable of diffracting and a light wave is.

d. capable of existing with a variety of frequencies and a light wave has a single frequency.

12. At the same temperature, sound waves have the fastest speed in:

a. rock

b. milk

c. oxygen

d. sand

13. Two sound waves are traveling through a container of nitrogen gas. The first wave has a wavelength of

1,5 m, while the second wave has a wavelength of 4,5 m. The velocity of the second wave must be:

a. 19 the velocity of the first wave.

b. 13 the velocity of the first wave.

c. the same as the velocity of the first wave.

d. three times larger than the velocity of the first wave.

e. nine times larger than the velocity of the first wave.

14. Sound travels at a speed of 340 m·s−1. A straw is 0,25 m long. The standing wave set up in such a straw

with one end closed has a wavelength of 1,0 m. The standing wave set up in such a straw with both ends

open has a wavelength of 0,50 m.

a. calculate the frequency of the sound created when you blow across the straw with the bottom end

closed.

b. calculate the frequency of the sound created when you blow across the straw with the bottom end open.

15. A lightning storm creates both lightning and thunder. You see the lightning almost immediately since light

travels at 3× 108m · s−1. After seeing the lightning, you count 5 s and then you hear the thunder. Calculate

the distance to the location of the storm.

16. A person is yelling from a second story window to another person standing at the garden gate, 50 m away.

If the speed of sound is 344 m·s−1, how long does it take the sound to reach the person standing at the

gate?

17. A piece of equipment has a warning label on it that says, "Caution! This instrument produces 140 decibels."

What safety precaution should you take before you turn on the instrument?

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CHAPTER 1. SOUND 1.150. EXERCISES

18. What property of sound is a measure of the amount of energy carried by a sound wave?

19. Person 1 speaks to person 2. Explain how the sound is created by person 1 and how it is possible for person

2 to hear the conversation.

20. Sound cannot travel in space. Discuss what other modes of communication astronauts can use when they

are outside the space shuttle?

21. An automatic focus camera uses an ultrasonic sound wave to focus on objects. The camera sends out

sound waves which are reflected off distant objects and return to the camera. A sensor detects the time it

takes for the waves to return and then determines the distance an object is from the camera. If a sound

wave (speed = 344 m·s−1) returns to the camera 0,150 s after leaving the camera, how far away is the

object?

22. Calculate the frequency (in Hz) and wavelength of the annoying sound made by a mosquito when it beats

its wings at the average rate of 600 wing beats per second. Assume the speed of the sound waves is

344 m·s−1.

23. How does halving the frequency of a wave source affect the speed of the waves?

24. Humans can detect frequencies as high as 20 000 Hz. Assuming the speed of sound in air is 344 m·s−1,

calculate the wavelength of the sound corresponding to the upper range of audible hearing.

25. An elephant trumpets at 10 Hz. Assuming the speed of sound in air is 344 m·s−1, calculate the wavelength

of this infrasonic sound wave made by the elephant.

26. A ship sends a signal out to determine the depth of the ocean. The signal returns 2,5 seconds later. If

sound travels at 1450 m.s−1 in sea water, how deep is the ocean at that point?

Find the answers with the shortcodes:

(1.) l4Y (2.) l41 (3.) l4C (4.) l4a (5.) l4x (6.) l4c

(7.) l4O (8.) l43 (9.) l4i (10.) l4l (11.) l4q (12.) lgh

(13.) lgS (14.) lgJ (15.) lgu (16.) lgz (17.) lgt (18.) lge

(19.) lgM (20.) lgL (21.) lgF (22.) lg6 (23.) lgH (24.) lgs

(25.) lgo (26.) lgA

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1.150. EXERCISES CHAPTER 1. SOUND

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Electromagnetic Radiation

1.151 Introduction

(section shortcode: P10063 )

This chapter will focus on the electromagnetic (EM) radiation. Electromagnetic radiation is a self-propagating

wave in space with electric and magnetic components. These components oscillate at right angles to each other

and to the direction of propagation, and are in phase with each other. Electromagnetic radiation is classified into

types according to the frequency of the wave: these types include, in order of increasing frequency, radio waves,

microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays.

1.152 Particle/Wave Nature of Electromagnetic Radiation

(section shortcode: P10064 )

If you watch a colony of ants walking up the wall, they look like a thin continuous black line. But as you look

closer, you see that the line is made up of thousands of separated black ants.

Light and all other types of electromagnetic radiation seems like a continuous wave at first, but when one performs

experiments with light, one can notice that light can have both wave and particle like properties. Just like the

individual ants, the light can also be made up of individual bundles of energy, or quanta of light.

Light has both wave-like and particle-like properties (wave–particle duality), but only shows one or the other,

depending on the kind of experiment we perform. A wave-type experiment shows the wave nature, and a particle-

type experiment shows particle nature. One cannot test the wave and the particle nature at the same time. A

particle of light is called a photon.

Definition: Photon

A photon is a quantum (energy packet) of light.

The particle nature of light can be demonstrated by the interaction of photons with matter. One way in which light

interacts with matter is via the photoelectric effect, which will be studied in detail in Chapter 20.

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1.153. THE WAVE NATURE OF ELECTROMAGNETIC RADIATIONCHAPTER 1. ELECTROMAGNETIC RADIATION

1.152.1 Particle/wave nature of electromagnetic radiation

1. Give examples of the behaviour of EM radiation which can best be explained using a wave model.

2. Give examples of the behaviour of EM radiation which can best be explained using a particle model.

Find the answers with the shortcodes:

(1.) l22 (2.) l2T

1.153 The wave nature of electromagnetic radiation

(section shortcode: P10065 )

Accelerating charges emit electromagnetic waves. We have seen that a changing electric field generates a mag-

netic field and a changing magnetic field generates an electric field. This is the principle behind the propagation

of electromagnetic waves, because electromagnetic waves, unlike sound waves, do not need a medium to travel

through. EM waves propagate when an electric field oscillating in one plane produces a magnetic field oscillat-

ing in a plane at right angles to it, which produces an oscillating electric field, and so on. The propagation of

electromagnetic waves can be described as mutual induction.

These mutually regenerating fields travel through empty space at a constant speed of 3×108m ·s−1, represented

by c.

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CHAPTER 1. ELECTROMAGNETIC RADIATION 1.154. ELECTROMAGNETIC SPECTRUM

1.154 Electromagnetic spectrum

(section shortcode: P10066 )

Figure 20.2: The electromagnetic spectrum as a function of frequency. The different types according to wave-

length are shown as well as everyday comparisons.

Observe the things around you, your friend sitting next to you, a large tree across the field. How is it that you are

able to see these things? What is it that is leaving your friend’s arm and entering your eye so that you can see his

arm? It is light. The light originally comes from the sun, or possibly a light bulb or burning fire. In physics, light is

given the more technical term electromagnetic radiation, which includes all forms of light, not just the form which

you can see with your eyes.

Electromagnetic radiation allows us to observe the world around us. It is this radiation which reflects off of the

objects around you and into your eye. The radiation your eye is sensitive to is only a small fraction of the total

radiation emitted in the physical universe. All of the different fractions taped together make up the electromagnetic

spectrum.

1.154.1 Dispersion

When white light is split into its component colours by a prism, you are looking at a portion of the electromagnetic

spectrum.

The wavelength of a particular electromagnetic radiation will depend on how it was created.

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1.154. ELECTROMAGNETIC SPECTRUM CHAPTER 1. ELECTROMAGNETIC RADIATION

1.154.2 Wave Nature of EM Radiation

1. List one source of electromagnetic waves. Hint: consider the spectrum diagram and look at the names we

give to different wavelengths.

2. Explain how an EM wave propagates, with the aid of a diagram.

3. What is the speed of light? What symbol is used to refer to the speed of light? Does the speed of light

change?

4. Do EM waves need a medium to travel through?

The radiation can take on any wavelength, which means that the spectrum is continuous. Physicists broke down

this continuous band into sections. Each section is defined by how the radiation is created, not the wavelength

of the radiation. But each category is continuous within the min and max wavelength of that category, meaning

there are no wavelengths excluded within some range.

The spectrum is in order of wavelength, with the shortest wavelength at one end and the longest wavelength at

the other. The spectrum is then broken down into categories as detailed in Table 20.1.

Category Range of Wavelengths (nm) Range of Frequencies (Hz)

gamma rays < 1 > 3× 1019

X-rays 1-10 3× 1017-3× 1019

ultraviolet light 10-400 7, 5× 1014-3× 1017

visible light 400-700 4, 3× 1014-7, 5× 1014

infrared 700-105 3× 1012-4, 3× 1019

microwave 105 − 108 3× 109-3× 1012

radio waves > 108 < 3× 109

Table 20.1: Electromagnetic spectrum

Since an electromagnetic wave is still a wave, the following equation that you learnt in Grade 10 still applies:

c = f · λ (20.1)

Exercise 20.1: EM spectrum I Calculate the frequency of red light with

a wavelength of 4, 2× 10−7 m

Solution to Exercise

Step 1. We use the formula: c = fλ to calculate frequency. The speed of

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CHAPTER 1. ELECTROMAGNETIC RADIATION 1.154. ELECTROMAGNETIC SPECTRUM

light is a constant 3× 108m/s.

c = fλ

3× 108 = f × 4, 2× 10−7

f = 7, 14× 1014Hz

(20.2)

Exercise 20.2: EM spectrum II Ultraviolet radiation has a wavelength

of 200 nm. What is the frequency of the radiation?

Solution to Exercise

Step 1. Recall that all radiation travels at the speed of light (c) in vacuum.

Since the question does not specify through what type of mate-

rial the Ultraviolet radiation is traveling, one can assume that it is

traveling through a vacuum. We can identify two properties of the

radiation - wavelength (200 nm) and speed (c).Step 2.

c = fλ

3× 108 = f × 200× 10−9

f = 1.5× 1015 Hz

(20.3)

Examples of some uses of electromagnetic waves are shown in Table 20.2.

Category Uses

gamma rays used to kill the bacteria in marshmallows and to sterilise medical equipment

X-rays used to image bone structures

ultraviolet light bees can see into the ultraviolet because flowers stand out more clearly at this frequency

visible light used by humans to observe the world

infrared night vision, heat sensors, laser metal cutting

microwave microwave ovens, radar

radio waves radio, television broadcasts

Table 20.2: Uses of EM waves

In theory the spectrum is infinite, although realistically we can only observe wavelengths from a few hundred

kilometers to those of gamma rays due to experimental limitations.

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1.155. THE PARTICLE NATURE OF ELECTROMAGNETIC RADIATIONCHAPTER 1. ELECTROMAGNETIC RADIATION

Humans experience electromagnetic waves differently depending on their wavelength. Our eyes are sensitive to

visible light while our skin is sensitive to infrared, and many wavelengths we do not detect at all.

Find the answers with the shortcodes:

(1.) l2c (2.) l2x (3.) l2a (4.) l2C

1.154.3 EM Radiation

1. Arrange the following types of EM radiation in order of increasing frequency: infrared, X-rays, ultraviolet,

visible, gamma.

2. Calculate the frequency of an EM wave with a wavelength of 400 nm.

3. Give an example of the use of each type of EM radiation, i.e. gamma rays, X-rays, ultraviolet light, visible

light, infrared, microwave and radio and TV waves.

Find the answers with the shortcodes:

(1.) l23 (2.) l2O (3.) l2i

1.155 The particle nature of electromagnetic radiation

(section shortcode: P10067 )

When we talk of electromagnetic radiation as a particle, we refer to photons, which are packets of energy. The

energy of the photon is related to the wavelength of electromagnetic radiation according to:

Definition: Planck’s constant

Planck’s constant is a physical constant named after Max Planck.

h = 6, 626× 10−34 J · s

The energy of a photon can be calculated using the formula: E = hf or E = h cλ

. Where E is the energy of the

photon in joules (J), h is planck’s constant, c is the speed of light, f is the frequency in hertz (Hz) and λ is the

wavelength in metres (m).

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CHAPTER 1. ELECTROMAGNETIC RADIATION1.155. THE PARTICLE NATURE OF ELECTROMAGNETIC RADIATION

Exercise 20.3: Calculating the energy of a photon I Calculate the

energy of a photon with a frequency of 3× 1018 Hz

Solution to Exercise

Step 1.

E = hf

= 6, 6× 10−34 × 3× 1018

= 2× 10−15 J

(20.4)

Exercise 20.4: Calculating the energy of a photon II What is the

energy of an ultraviolet photon with a wavelength of 200 nm?

Solution to Exercise

Step 1. We are required to calculate the energy associated with a photon

of ultraviolet light with a wavelength of 200 nm.

We can use:

E = hc

λ(20.5)

Step 2.

E = h cλ

=(

6, 626× 10−34)

3×108

200×10−9

= 9, 939× 10−10 J

(20.6)

1.155.1 Exercise - particle nature of EM waves

1. How is the energy of a photon related to its frequency and wavelength?

2. Calculate the energy of a photon of EM radiation with a frequency of 1012 Hz.

3. Determine the energy of a photon of EM radiation with a wavelength of 600 nm.

Find the answers with the shortcodes:

(1.) l2H (2.) l26 (3.) l2F

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1.156. PENETRATING ABILITY OF ELECTROMAGNETIC RADIATIONCHAPTER 1. ELECTROMAGNETIC RADIATION

1.156 Penetrating ability of electromagnetic radiation

(section shortcode: P10068 )

Different kinds of electromagnetic radiation have different penetrabilities. For example, if we take the human body

as the object. Infrared light is emitted by the human body. Visible light is reflected off the surface of the human

body, ultra-violet light (from sunlight) damages the skin, but X-rays are able to penetrate the skin and bone and

allow for pictures of the inside of the human body to be taken.

If we compare the energy of visible light to the energy of X-rays, we find that X-rays have a much higher energy.

Usually, kinds of electromagnetic radiation with higher energy have higher penetrabilities than those with low

energies.

Certain kinds of electromagnetic radiation such as ultra-violet radiation, X-rays and gamma rays are very dan-

gerous. Radiation such as these are called ionising radiation. Ionising radiation transfers energy as it passes

through matter, breaking molecular bonds and creating ions.

Excessive exposure to radiation, including sunlight, X-rays and all nuclear radiation, can cause destruction of

biological tissue.

1.156.1 Ultraviolet(UV) radiation and the skin

UVA and UVB are different ranges of frequencies for ultraviolet (UV) light. UVA and UVB can damage collagen

fibres which results in the speeding up skin aging. In general, UVA is the least harmful, but it can contribute

to the aging of skin, DNA damage and possibly skin cancer. It penetrates deeply and does not cause sunburn.

Because it does not cause reddening of the skin (erythema) it cannot be measured in the SPF testing. There is

no good clinical measurement of the blocking of UVA radiation, but it is important that sunscreen block both UVA

and UVB.

UVB light can cause skin cancer. The radiation excites DNA molecules in skin cells, resulting in possible muta-

tions, which can cause cancer. This cancer connection is one reason for concern about ozone depletion and the

ozone hole.

As a defense against UV radiation, the body tans when exposed to moderate (depending on skin type) levels

of radiation by releasing the brown pigment melanin. This helps to block UV penetration and prevent damage

to the vulnerable skin tissues deeper down. Suntan lotion, often referred to as sunblock or sunscreen, partly

blocks UV and is widely available. Most of these products contain an SPF rating that describes the amount of

protection given. This protection, however, applies only to UVB rays responsible for sunburn and not to UVA rays

that penetrate more deeply into the skin and may also be responsible for causing cancer and wrinkles. Some

sunscreen lotion now includes compounds such as titanium dioxide which helps protect against UVA rays. Other

UVA blocking compounds found in sunscreen include zinc oxide and avobenzone.

What makes a good sunscreen?

• UVB protection: Padimate O, Homosalate, Octisalate (octyl salicylate), Octinoxate (octyl methoxycinna-

mate)

• UVA protection: Avobenzone

• UVA/UVB protection: Octocrylene, titanium dioxide, zinc oxide, Mexoryl (ecamsule)

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CHAPTER 1. ELECTROMAGNETIC RADIATION1.156. PENETRATING ABILITY OF ELECTROMAGNETIC RADIATION

Another means to block UV is by wearing sun protective clothing. This is clothing that has a UPF rating that

describes the protection given against both UVA and UVB.

1.156.2 Ultraviolet radiation and the eyes

High intensities of UVB light are hazardous to the eyes, and exposure can cause welder’s flash (photo keratitis or

arc eye) and may lead to cataracts, pterygium and pinguecula formation.

Protective eyewear is beneficial to those who are working with or those who might be exposed to ultraviolet radi-

ation, particularly short wave UV. Given that light may reach the eye from the sides, full coverage eye protection

is usually warranted if there is an increased risk of exposure, as in high altitude mountaineering. Mountaineers

are exposed to higher than ordinary levels of UV radiation, both because there is less atmospheric filtering and

because of reflection from snow and ice.

Ordinary, untreated eyeglasses give some protection. Most plastic lenses give more protection than glass lenses.

Some plastic lens materials, such as polycarbonate, block most UV. There are protective treatments available for

eyeglass lenses that need it which will give better protection. But even a treatment that completely blocks UV will

not protect the eye from light that arrives around the lens. To convince yourself of the potential dangers of stray

UV light, cover your lenses with something opaque, like aluminum foil, stand next to a bright light, and consider

how much light you see, despite the complete blockage of the lenses. Most contact lenses help to protect the

retina by absorbing UV radiation.

1.156.3 X-rays

While x-rays are used significantly in medicine, prolonged exposure to X-rays can lead to cell damage and cancer.

For example, a mammogram is an x-ray of the human breast to detect breast cancer, but if a woman starts having

regular mammograms when she is too young, her chances of getting breast cancer increases.

1.156.4 Gamma-rays

Due to the high energy of gamma-rays, they are able to cause serious damage when absorbed by living cells.

Gamma-rays are not stopped by the skin and can induce DNA alteration by interfering with the genetic material of

the cell. DNA double-strand breaks are generally accepted to be the most biologically significant lesion by which

ionising radiation causes cancer and hereditary disease.

A study done on Russian nuclear workers exposed to external whole-body gamma-radiation at high cumulative

doses shows a link between radiation exposure and death from leukaemia, lung, liver, skeletal and other solid

cancers.

Cellphones and electromagnetic radiation

Cellphone radiation and health concerns have been raised, especially following the enormous increase in the use

of wireless mobile telephony throughout the world. This is because mobile phones use electromagnetic waves in

the microwave range. These concerns have induced a large body of research. Concerns about effects on health

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1.157. SUMMARY CHAPTER 1. ELECTROMAGNETIC RADIATION

have also been raised regarding other digital wireless systems, such as data communication networks. In 2009

the World Health Organisation announced that they have found a link between brain cancer and cellphones.

Cellphone users are recommended to minimise radiation, by for example:

1. Use hands-free to decrease the radiation to the head.

2. Keep the mobile phone away from the body.

3. Do not telephone in a car without an external antenna.

1.156.5 Exercise - Penetrating ability of EM radiation

1. Indicate the penetrating ability of the different kinds of EM radiation and relate it to energy of the radiation.

2. Describe the dangers of gamma rays, X-rays and the damaging effect of ultra-violet radiation on skin.

Find the answers with the shortcodes:

(1.) l2l (2.) l2q

1.157 Summary

(section shortcode: P10069 )

1. Electromagnetic radiation has both a wave and a particle nature.

2. Electromagnetic waves travel at a speed of 3× 108 m · s−1 in a vaccum.

3. The Electromagnetic spectrum consists of the follwing types of radiation: radio waves, microwaves, infrared,

visible, ultraviolet, X-rays, gamma-rays.

4. Gamma-rays have the most energy and are the most penetrating, while radio waves have the lowest energy

and are the least penetrating.

1.158 End of chapter exercise

(section shortcode: P10070 )

1. What is the energy of a photon of EM radiation with a frequency of 3× 108 Hz?

2. What is the energy of a photon of light with a wavelength of 660 nm?

3. List the main types of electromagnetic radiation in order of increasing wavelength.

4. List the main uses of:

a. radio waves

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CHAPTER 1. ELECTROMAGNETIC RADIATION 1.158. END OF CHAPTER EXERCISE

b. infrared

c. gamma rays

d. X-rays

5. Explain why we need to protect ourselves from ultraviolet radiation from the Sun.

6. List some advantages and disadvantages of using X-rays.

7. What precautions should we take when using cell phones?

8. Write a short essay on a type of electromagnetic waves. You should look at uses, advantages and disad-

vantages of your chosen radiation.

9. Explain why some types of electromagnetic radiation are more penetrating than others.

Find the answers with the shortcodes:

(1.) l4J (2.) l4u (3.) l2r (4.) l21 (5.) l2Y (6.) l4h

(7.) l4S (8.) l24 (9.) l2g

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1.158. END OF CHAPTER EXERCISE CHAPTER 1. ELECTROMAGNETIC RADIATION

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Magnetism

1.159 Introduction

(section shortcode: P10071 )

Magnetism is a force that certain kinds of objects, which are called ‘magnetic’ objects, can exert on each other

without physically touching. A magnetic object is surrounded by a magnetic ‘field’ that gets weaker as one moves

further away from the object. A second object can feel a magnetic force from the first object because it feels the

magnetic field of the first object.

Humans have known about magnetism for many thousands of years. For example, lodestone is a magnetised

form of the iron oxide mineral magnetite. It has the property of attracting iron objects. It is referred to in old

European and Asian historical records; from around 800 BCE in Europe and around 2 600 BCE in Asia.

NOTE: The root of the English word magnet is from the Greek word magnes, probably

from Magnesia in Asia Minor, once an important source of lodestone.

1.160 Magnetic fields

(section shortcode: P10072 )

A magnetic field is a region in space where a magnet or object made of magnetic material will experience a

non-contact force.

Electrons inside any object have magnetic fields associated with them. In most materials these fields point in

all directions, so the net magnetic field is zero. For example, in the plastic ball below, the directions of the

magnetic fields of the electrons (shown by the arrows) are pointing in different directions and cancel each other

out. Therefore the plastic ball is not magnetic and has no magnetic field.

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1.160. MAGNETIC FIELDS CHAPTER 1. MAGNETISM

In some materials (e.g. iron), called ferromagnetic materials, there are regions called domains, where the

electrons’ magnetic fields line up with each other. All the atoms in each domain are grouped together so that the

magnetic fields from their electrons point the same way. The picture shows a piece of an iron needle zoomed in

to show the domains with the electric fields lined up inside them.

In permanent magnets, many domains are lined up, resulting in a net magnetic field. Objects made from ferro-

magnetic materials can be magnetised, for example by rubbing a magnet along the object in one direction. This

causes the magnetic fields of most, or all, of the domains to line up in one direction. As a result the object as a

whole will have a net magnetic field. It is magnetic. Once a ferromagnetic object has been magnetised, it can

stay magnetic without another magnet being nearby (i.e. without being in another magnetic field). In the picture

below, the needle has been magnetised because the magnetic fields in all the domains are pointing in the same

direction.

1.160.1 Investigation : Ferromagnetic materials and magnetisation

1. Find 2 paper clips. Put the paper clips close together and observe what happens.

a. What happens to the paper clips?

b. Are the paper clips magnetic?

2. Now take a permanent bar magnet and rub it once along 1 of the paper clips. Remove the magnet and put

the paper clip which was touched by the magnet close to the other paper clip and observe what happens.

Does the untouched paper clip feel a force on it? If so, is the force attractive or repulsive?

3. Rub the same paper clip a few more times with the bar magnet, in the same direction as before. Put the

paper clip close to the other one and observe what happens.

a. Is there any difference to what happened in step 2?

b. If there is a difference, what is the reason for it?

c. Is the paper clip which was rubbed repeatedly by the magnet now magnetised?

d. What is the difference between the two paper clips at the level of their atoms and electrons?

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CHAPTER 1. MAGNETISM 1.161. PERMANENT MAGNETS

4. Now, find a metal knitting needle, or a metal ruler, or other metal object. Rub the bar magnet along the

knitting needle a few times in the same direction. Now put the knitting needle close to the paper clips and

observe what happens.

a. Does the knitting needle attract the paper clips?

b. What does this tell you about the material of the knitting needle? Is it ferromagnetic?

5. Repeat this experiment with objects made from other materials. Which materials appear to be ferromagnetic

and which are not? Put your answers in a table.

A ferromagnetic material is a substance that shows spontaneous magnetisation.

1.161 Permanent magnets

(section shortcode: P10073 )

1.161.1 The poles of permanent magnets

Because the domains in a permanent magnet all line up in a particular direction, the magnet has a pair of opposite

poles, called north (usually shortened to N) and south (usually shortened to S). Even if the magnet is cut into

tiny pieces, each piece will still have both a N and a S pole. These magnetic poles always occur in pairs. In

nature, we never find a north magnetic pole or south magnetic pole on its own.

Magnetic fields are different from gravitational and electric fields. In nature, positive and negative electric charges

can be found on their own, but you never find just a north magnetic pole or south magnetic pole on its own. On

the very small scale, zooming in to the size of atoms, magnetic fields are caused by moving charges (i.e. the

negatively charged electrons).

1.161.2 Magnetic attraction and repulsion

Like (identical) poles of magnets repel one another whilst unlike (opposite) poles attract. This means that two

N poles or two S poles will push away from each other while a N pole and a S pole will be drawn towards each

other.

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1.161. PERMANENT MAGNETS CHAPTER 1. MAGNETISM

Definition: Attraction and Repulsion

Like poles of magnets repel each other whilst unlike poles attract each other.

Exercise 21.1: Attraction and Repulsion Do you think the following

magnets will repel or be attracted to each other?

Solution to Exercise

Step 1. We are required to determine whether the two magnets will repel

each other or be attracted to each other.

Step 2. We are given two magnets with the N pole of one approaching the

N pole of the other.

Step 3. Since both poles are the same, the magnets will repel each other.

Exercise 21.2: Attraction and repulsion Do you think the following

magnets will repel or be attracted to each other?

Solution to Exercise

Step 1. We are required to determine whether the two magnets will repel

each other or be attracted to each other.

Step 2. We are given two magnets with the N pole of one approaching the

S pole of the other.

Step 3. Since both poles are the different, the magnets will be attracted to

each other.

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1.161.3 Representing magnetic fields

Magnetic fields can be represented using magnetic field lines starting at the North pole and ending at the

South pole. Although the magnetic field of a permanent magnet is everywhere surrounding the magnet (in all

three dimensions), we draw only some of the field lines to represent the field (usually only a two-dimensional

cross-section is shown in drawings).

In areas where the magnetic field is strong, the field lines are closer together. Where the field is weaker, the

field lines are drawn further apart. The number of field lines drawn crossing a given two-dimensional surface is

referred to as the magnetic flux. The magnetic flux is used as a measure of the strength of the magnetic field

over that surface.

TIP:

1.Field lines never cross.

2.Arrows drawn on the field lines indicate the direction of the field.

3.A magnetic field points from the north to the south pole of a magnet.

Investigation : Field around a Bar Magnet

Take a bar magnet and place it on a flat surface. Place a sheet of white paper over the bar magnet and sprinkle

some iron filings onto the paper. Give the paper a shake to evenly distribute the iron filings. In your workbook,

draw the bar magnet and the pattern formed by the iron filings. Draw the pattern formed when you rotate the bar

magnet to a different angle as shown below.

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1.161. PERMANENT MAGNETS CHAPTER 1. MAGNETISM

As the activity shows, one can map the magnetic field of a magnet by placing it underneath a piece of paper and

sprinkling iron filings on top. The iron filings line themselves up parallel to the magnetic field.

Investigation : Field around a Pair of Bar Magnets

Take two bar magnets and place them a short distance apart such that they are repelling each other. Place a

sheet of white paper over the bar magnets and sprinkle some iron filings onto the paper. Give the paper a shake

to evenly distribute the iron filings. In your workbook, draw both the bar magnets and the pattern formed by the

iron filings. Repeat the procedure for two bar magnets attracting each other and draw what the pattern looks

like for this situation. Make a note of the shape of the lines formed by the iron filings, as well as their size and

their direction for both arrangements of the bar magnet. What does the pattern look like when you place both bar

magnets side by side?

As already said, opposite poles of a magnet attract each other and bringing them together causes their magnetic

field lines to converge (come together). Like poles of a magnet repel each other and bringing them together

causes their magnetic field lines to diverge (bend out from each other).

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CHAPTER 1. MAGNETISM 1.162. THE COMPASS AND THE EARTH’S MAGNETIC FIELD

Ferromagnetism and Retentivity

Ferromagnetism is a phenomenon shown by materials like iron, nickel or cobalt. These materials can form

permanent magnets. They always magnetise so as to be attracted to a magnet, no matter which magnetic pole

is brought toward the unmagnetised iron/nickel/cobalt.

The ability of a ferromagnetic material to retain its magnetisation after an external field is removed is called its

retentivity.

Paramagnetic materials are materials like aluminium or platinum, which become magnetised in an external

magnetic field in a similar way to ferromagnetic materials. However, they lose their magnetism when the external

magnetic field is removed.

Diamagnetism is shown by materials like copper or bismuth, which become magnetised in a magnetic field with

a polarity opposite to the external magnetic field. Unlike iron, they are slightly repelled by a magnet.

1.162 The compass and the earth’s magnetic field

(section shortcode: P10074 )

A compass is an instrument which is used to find the direction of a magnetic field. A compass consists of a

small metal needle which is magnetised itself and which is free to turn in any direction. Therefore, when in the

presence of a magnetic field, the needle is able to line up in the same direction as the field.

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1.162. THE COMPASS AND THE EARTH’S MAGNETIC FIELD CHAPTER 1. MAGNETISM

NOTE: Lodestone, a magnetised form of iron-oxide, was found to orientate itself in

a north-south direction if left free to rotate by suspension on a string or on a float in

water. Lodestone was therefore used as an early navigational compass.

Compasses are mainly used in navigation to find direction on the earth. This works because the earth itself has

a magnetic field which is similar to that of a bar magnet (see the picture below). The compass needle aligns with

the earth’s magnetic field direction and points north-south. Once you know where north is, you can figure out any

other direction. A picture of a compass is shown below:

Some animals can detect magnetic fields, which helps them orientate themselves and navigate. Animals which

can do this include pigeons, bees, Monarch butterflies, sea turtles and certain fish.

1.162.1 The earth’s magnetic field

In the picture below, you can see a representation of the earth’s magnetic field which is very similar to the

magnetic field of a giant bar magnet like the one on the right of the picture. So the earth has two sets of north

poles and south poles: geographic poles and magnetic poles.

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CHAPTER 1. MAGNETISM 1.162. THE COMPASS AND THE EARTH’S MAGNETIC FIELD

The earth’s magnetic field is thought to be caused by flowing liquid metals in the outer core which causes electric

currents and a magnetic field. From the picture you can see that the direction of magnetic north and true north

are not identical. The geographic north pole, which is the point through which the earth’s rotation axis goes, is

about 11,5o away from the direction of the magnetic north pole (which is where a compass will point). However,

the magnetic poles shift slightly all the time.

Another interesting thing to note is that if we think of the earth as a big bar magnet, and we know that magnetic

field lines always point from north to south, then the compass tells us that what we call the magnetic north pole

is actually the south pole of the bar magnet!

NOTE: The direction of the earth’s magnetic field flips direction about once every

200 000 years! You can picture this as a bar magnet whose north and south pole

periodically switch sides. The reason for this is still not fully understood.

Phenomena related to the Earth’s magnetic field

The importance of the magnetic field to life on Earth

The earth’s magnetic field is very important for humans and other animals on earth because it protects us from

being bombarded (hit) by high energy charged particles which are emitted by the Sun. The stream of charged

particles (mainly positively charged protons and negatively charged electrons) coming from the sun is called the

solar wind. When these particles come close to the Earth, they become trapped by the Earth’s magnetic field

and cannot shower down to the surface where they can harm living organisms. Astronauts in space are at risk of

being irradiated by the solar wind because they are outside the zones where the charged particles are trapped.

The region above Earth’s atmosphere in which charged particles are affected by Earth’s magnetic field is called

the magnetosphere. Relatively often, in addition to the usual solar wind, the Sun may eject a large bubble of

material (protons and electrons) with its own magnetic field from its outer atmosphere. Sometimes these bubbles

travel towards the Earth where their magnetic fields can join with Earth’s magnetic field. When this happens a

huge amount of energy is released into the Earth’s magnetosphere, causing a geomagnetic storm. These storms

cause rapid changes in the Earth’s magnetosphere which in turn may affect electric and magnetic systems on

the Earth such as power grids, cellphone networks, and other electronic systems.

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1.163. SUMMARY CHAPTER 1. MAGNETISM

Aurorae (pronounced Or-roar-ee)

Another effect caused by the Earth’s magnetic field is the spectacular Northern and Southern Lights, which are

also called the Aurora Borealis and the Aurora Australis respectively. When charged particles from the solar wind

reach the Earth’s magnetosphere, they spiral along the magnetic field lines towards the North and South poles.

If they collide with particles in the Earth’s atmosphere, they can cause red or green lights which stretch across

a large part of the sky and which is called the aurora. Because this only happens close to the North and South

poles, we cannot see the aurorae from South Africa. However, people living in the high Northern latitudes in

Canada, Sweden, and Finland, for example, often see the Northern lights.

This simulation shows you the Earth’s magnetic field and a compass. (Simulation: lb9)

1.163 Summary

(section shortcode: P10075 )

1. Magnets have two poles - North and South.

2. Some substances can be easily magnetised.

3. Like poles repel each other and unlike poles attract each other.

4. The Earth also has a magnetic field.

5. A compass can be used to find the magnetic north pole and help us find our direction.

This video provides a summary of the work covered in this chapter.

Khan academy video on magnets (Video: EZURLhttp://www.youtube.com/v/8Y4JSp5U82I&rel=0 )

1.164 End of chapter exercises

(section shortcode: P10077 )

1. Describe what is meant by the term magnetic field.

2. Use words and pictures to explain why permanent magnets have a magnetic field around them. Refer to

domains in your explanation.

3. What is a magnet?

4. What happens to the poles of a magnet if it is cut into pieces?

5. What happens when like magnetic poles are brought close together?

6. What happens when unlike magnetic poles are brought close together?

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CHAPTER 1. MAGNETISM 1.164. END OF CHAPTER EXERCISES

7. Draw the shape of the magnetic field around a bar magnet.

8. Explain how a compass indicates the direction of a magnetic field.

9. Compare the magnetic field of the Earth to the magnetic field of a bar magnet using words and diagrams.

10. Explain the difference between the geographical north pole and the magnetic north pole of the Earth.

11. Give examples of phenomena that are affected by Earth’s magnetic field.

12. Draw a diagram showing the magnetic field around the Earth.

Find the answers with the shortcodes:

(1.) llS (2.) lia (3.) lix (4.) lic (5.) liO (6.) li3

(7.) lii (8.) llu (9.) lil (10.) llh (11.) llJ (12.) llt

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Electrostatics

1.165 Introduction

(section shortcode: P10078 )

Electrostatics is the study of electric charge which is static (not moving). In this chapter we will look at some of

the basic principle of electric charge as well as the principle of conservation of charge.

1.166 Two kinds of charge

(section shortcode: P10079 )

All objects surrounding us (including people!) contain large amounts of electric charge. There are two types of

electric charge: positive charge and negative charge. If the same amounts of negative and positive charge

are brought together, they neutralise each other and there is no net charge. Neutral objects are objects which

contain equal amouts of positive and negative charges. However, if there is a little bit more of one type of charge

than the other on the object then the object is said to be electrically charged. The picture below shows what the

distribution of charges might look like for a neutral, positively charged and negatively charged object.

1.167 Unit of charge

(section shortcode: P10080 )

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1.168. CONSERVATION OF CHARGE CHAPTER 1. ELECTROSTATICS

Charge is measured in units called coulombs (C). A coulomb of charge is a very large charge. In electrostatics

we therefore often work with charge in microcoulombs (1 µC = 1 × 10−6 C) and nanocoulombs (1 nC = 1 ×10−9 C).

1.168 Conservation of charge

(section shortcode: P10081 )

Objects may become charged in many ways, including by contact with or being rubbed by other objects. This

means that they can gain extra negative or positive charge. For example, charging happens when you rub your

feet against the carpet. When you then touch something metallic or another person, you feel a shock as the

excess charge that you have collected is discharged.

TIP: Charge, like energy, cannot be created or destroyed. We say that charge is

conserved.

The principle of conservation of charge states that the net charge of an isolated system remains constant during

any physical process, e.g. when two charges make contact and are separated again.

When you rub your feet against the carpet, negative charge is transferred to you from the carpet. The carpet will

then become positively charged by the same amount.

Another example is to take two neutral objects such as a plastic ruler and a cotton cloth (handkerchief). To begin,

the two objects are neutral (i.e. have the same amounts of positive and negative charge).

Now, if the cotton cloth is used to rub the ruler, negative charge is transferred from the cloth to the ruler. The ruler

is now negatively charged (i.e. has an excess of electrons) and the cloth is positively charged (i.e. is electron

deficient). If you count up all the positive and negative charges at the beginning and the end, there are still the

same amount. i.e. total charge has been conserved !

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CHAPTER 1. ELECTROSTATICS 1.169. FORCE BETWEEN CHARGES

Note that in this example the numbers are made up to be easy to calculate. In the real world only a tiny fraction

of the charges would move from one object to the other, but the total charge would still be conserved.

The following simulation will help you understand what happens when you rub an object against another object.

(Simulation: lbX)

NOTE: The process of materials becoming charged when they come into contact with

other materials is known as tribo-electric charging. Materials can be arranged in a

tribo-electric series according to whether they are more positive or more negative.

This tribo-electric series can allow us to determine whether one material is likely to

become charged from another material. For example, amber is more negative than

wool and so if a piece of wool is rubbed against a piece of amber then the amber will

become negatively charged.

1.169 Force between Charges

(section shortcode: P10082 )

The force exerted by non-moving (static) charges on each other is called the electrostatic force. The electro-

static force between:

• like charges is repulsive

• opposite (unlike) charges is attractive.

In other words, like charges repel each other while opposite charges attract each other. This is different to the

gravitational force which is only attractive.

The closer together the charges are, the stronger the electrostatic force between them.

1.169.1 Experiment : Electrostatic Force

You can easily test that like charges repel and unlike charges attract each other by doing a very simple experiment.

Take a glass rod and rub it with a piece of silk, then hang it from its middle with a piece string so that it is free to

move. If you then bring another glass rod which you have also charged in the same way next to it, you will see

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1.169. FORCE BETWEEN CHARGES CHAPTER 1. ELECTROSTATICS

the rod on the string turn away from the rod in your hand i.e. it is repelled. If, however, you take a plastic rod, rub

it with a piece of fur and then bring it close to the rod on the string, you will see the rod on the string turn towards

the rod in your hand i.e. it is attracted.

This happens because when you rub the glass with silk, tiny amounts of negative charge are transferred from

the glass onto the silk, which causes the glass to have less negative charge than positive charge, making it

positively charged. When you rub the plastic rod with the fur, you transfer tiny amounts of negative charge onto

the rod and so it has more negative charge than positive charge on it, making it negatively charged.

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CHAPTER 1. ELECTROSTATICS 1.169. FORCE BETWEEN CHARGES

Exercise 22.1: Application of electrostatic forces Two charged metal

spheres hang from strings and are free to move as shown in the picture

below. The right hand sphere is positively charged. The charge on the

left hand sphere is unknown.

The left sphere is now brought close to the right sphere.

1. If the left hand sphere swings towards the right hand sphere, what

can you say about the charge on the left sphere and why?

2. If the left hand sphere swings away from the right hand sphere,

what can you say about the charge on the left sphere and why?

Solution to Exercise

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1.170. CONDUCTORS AND INSULATORS CHAPTER 1. ELECTROSTATICS

Step 1. In the first case, we have a sphere with positive charge which is

attracting the left charged sphere. We need to find the charge on

the left sphere.

Step 2. We are dealing with electrostatic forces between charged objects.

Therefore, we know that like charges repel each other and opposite

charges attract each other.

Step 3. a. In the first case, the positively charged sphere is attracting

the left sphere. Since an electrostatic force between un-

like charges is attractive, the left sphere must be negatively

charged.

b. In the second case, the positively charged sphere repels the

left sphere. Like charges repel each other. Therefore, the left

sphere must now also be positively charged.

NOTE: The word ’electron’ comes from the Greek word for amber. The ancient Greeks

observed that if you rubbed a piece of amber, you could use it to pick up bits of straw.

1.170 Conductors and insulators

(section shortcode: P10083 )

All atoms are electrically neutral i.e. they have the same amounts of negative and positive charge inside them. By

convention, the electrons carry negative charge and the protons carry positive charge. The basic unit of charge,

called the elementary charge, e, is the amount of charge carried by one electron.

The charge on a single electron is qe = 1, 6x10−19 C. All other charges in the universe consist of an interger

multiple of this charge (i.e. Q = nqe). This is known as charge quantisation.

All the matter and materials on earth are made up of atoms. Some materials allow electrons to move relatively

freely through them (e.g. most metals, the human body). These materials are called conductors.

Other materials do not allow the charge carriers, the electrons, to move through them (e.g. plastic, glass). The

electrons are bound to the atoms in the material. These materials are called non-conductors or insulators.

If an excess of charge is placed on an insulator, it will stay where it is put and there will be a concentration of

charge in that area of the object. However, if an excess of charge is placed on a conductor, the like charges will

repel each other and spread out over the outside surface of the object. When two conductors are made to touch,

the total charge on them is shared between the two. If the two conductors are identical, then each conductor will

be left with half of the total charge.

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CHAPTER 1. ELECTROSTATICS 1.170. CONDUCTORS AND INSULATORS

NOTE: The electrostatic force determines the arrangement of charge on the surface of

conductors. This is possible because charges can move inside a conductive material.

When we place a charge on a spherical conductor the repulsive forces between the

individual like charges cause them to spread uniformly over the surface of the sphere.

However, for conductors with non-regular shapes, there is a concentration of charge

near the point or points of the object. Notice in Figure 22.9 that we show a concen-

tration of charge with more − or + signs, while we represent uniformly spread charges

with uniformly spaced − or + signs.

This collection of charge can actually allow charge to leak off the conductor if the point

is sharp enough. It is for this reason that buildings often have a lightning rod on the

roof to remove any charge the building has collected. This minimises the possibility of

the building being struck by lightning. This “spreading out” of charge would not occur

if we were to place the charge on an insulator since charge cannot move in insulators.

ASIDE: The basic unit of charge, namely the elementary charge is carried by the elec-

tron (equal to 1.602×10−19 C!). In a conducting material (e.g. copper), when the

atoms bond to form the material, some of the outermost, loosely bound electrons be-

come detached from the individual atoms and so become free to move around. The

charge carried by these electrons can move around in the material. In insulators, there

are very few, if any, free electrons and so the charge cannot move around in the mate-

rial.

NOTE: In 1909 Robert Millikan and Harvey Fletcher measured the charge on an elec-

tron. This experiment is now known as Millikan’s oil drop experiment. Millikan and

Fletcher sprayed oil droplets into the space between two charged plates and used what

they knew about forces and in particular the electric force to determine the charge on

an electron.

Exercise 22.2: Conducting spheres and movement of charge I have

2 charged metal conducting spheres which are identical except for hav-

ing different charge. Sphere A has a charge of -5 nC and sphere B has

a charge of -3 nC. I then bring the spheres together so that they touch

each other. Afterwards I move the two spheres apart so that they are no

longer touching.

1. What happens to the charge on the two spheres?

2. What is the final charge on each sphere?

Solution to Exercise

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1.170. CONDUCTORS AND INSULATORS CHAPTER 1. ELECTROSTATICS

Step 1. We have two identical negatively charged conducting spheres

which are brought together to touch each other and then taken

apart again. We need to explain what happens to the charge on

each sphere and what the final charge on each sphere is after they

are moved apart.

Step 2. We know that the charge carriers in conductors are free to move

around and that charge on a conductor spreads itself out on the

surface of the conductor.

Step 3. a. When the two conducting spheres are brought together to

touch, it is as though they become one single big conductor

and the total charge of the two spheres spreads out across the

whole surface of the touching spheres. When the spheres are

moved apart again, each one is left with half of the total original

charge.

b. Before the spheres touch, the total charge is: -5 nC + (-3) nC

= -8 nC. When they touch they share out the -8 nC across their

whole surface. When they are removed from each other, each

is left with half of the original charge:

−8 nC /2 = −4 nC (22.1)

on each sphere.

In the previous example we worked out what happens when two identical conductors are allowed to touch. We

noticed that if we take two identically sized conducting spheres on insulating stands and bring them together so

that they touch, each sphere will have the same final charge. If the initial charge on the first sphere is Q1 and the

initial charge on the second sphere is Q2, then the final charge on the two spheres after they have been brought

into contact is:

Q =Q1 +Q2

2(22.2)

1.170.1 The electroscope

The electroscope is a very sensitive instrument which can be used to detect electric charge. A diagram of a

gold leaf electroscope is shown the figure below. The electroscope consists of a glass container with a metal rod

inside which has 2 thin pieces of gold foil attached. The other end of the metal rod has a metal plate attached to

it outside the glass container.

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The electroscope detects charge in the following way: A charged object, like the positively charged rod in the

picture, is brought close to (but not touching) the neutral metal plate of the electroscope. This causes negative

charge in the gold foil, metal rod, and metal plate, to be attracted to the positive rod. Because the metal (gold is

a metal too!) is a conductor, the charge can move freely from the foil up the metal rod and onto the metal plate.

There is now more negative charge on the plate and more positive charge on the gold foil leaves. This is called

inducing a charge on the metal plate. It is important to remember that the electroscope is still neutral (the total

positive and negative charges are the same), the charges have just been induced to move to different parts of

the instrument! The induced positive charge on the gold leaves forces them apart since like charges repel! This

is how we can tell that the rod is charged. If the rod is now moved away from the metal plate, the charge in the

electroscope will spread itself out evenly again and the leaves will fall down because there will no longer be an

induced charge on them.

Grounding

If you were to bring the charged rod close to the uncharged electroscope, and then you touched the metal plate

with your finger at the same time, this would cause charge to flow up from the ground (the earth), through your

body onto the metal plate. Connecting to the earth so charge flows is called grounding. The charge flowing

onto the plate is opposite to the charge on the rod, since it is attracted to the charge on the rod. Therefore,

for our picture, the charge flowing onto the plate would be negative. Now that charge has been added to the

electroscope, it is no longer neutral, but has an excess of negative charge. Now if we move the rod away, the

leaves will remain apart because they have an excess of negative charge and they repel each other. If we ground

the electroscope again (this time without the charged rod nearby), the excess charge will flow back into the earth,

leaving it neutral.

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1.171. ATTRACTION BETWEEN CHARGED AND UNCHARGED OBJECTSCHAPTER 1. ELECTROSTATICS

1.171 Attraction between charged and uncharged objects

(section shortcode: P10084 )

1.171.1 Polarisation of Insulators

Unlike conductors, the electrons in insulators (non-conductors) are bound to the atoms of the insulator and cannot

move around freely through the material. However, a charged object can still exert a force on a neutral insulator

due to a phenomenon called polarisation.

If a positively charged rod is brought close to a neutral insulator such as polystyrene, it can attract the bound

electrons to move round to the side of the atoms which is closest to the rod and cause the positive nuclei to

move slightly to the opposite side of the atoms. This process is called polarisation. Although it is a very small

(microscopic) effect, if there are many atoms and the polarised object is light (e.g. a small polystyrene ball), it can

add up to enough force to cause the object to be attracted onto the charged rod. Remember, that the polystyrene

is only polarised, not charged. The polystyrene ball is still neutral since no charge was added or removed from

it. The picture shows a not-to-scale view of the polarised atoms in the polystyrene ball:

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CHAPTER 1. ELECTROSTATICS 1.172. SUMMARY

Some materials are made up of molecules which are already polarised. These are molecules which have a more

positive and a more negative side but are still neutral overall. Just as a polarised polystyrene ball can be attracted

to a charged rod, these materials are also affected if brought close to a charged object.

Water is an example of a substance which is made of polarised molecules. If a positively charged rod is brought

close to a stream of water, the molecules can rotate so that the negative sides all line up towards the rod. The

stream of water will then be attracted to the rod since opposite charges attract.

1.172 Summary

(section shortcode: P10085 )

1. Objects can be positively charged, negatively charged or neutral.

2. Objects that are neutral have equal numbers of positive and negative charge.

3. Unlike charges are attracted to each other and like charges are repelled from each other.

4. Charge is neither created nor destroyed, it can only be transferred.

5. Charge is measured in coulombs (C).

6. Conductors allow charge to move through them easily.

7. Insulators do not allow charge to move through them easily.

The following presentation is a summary of the work covered in this chapter. Note that the last two slides are not

needed for exam purposes, but are included for general interest.

(Presentation: P10086 )

1.173 End of chapter exercise

(section shortcode: P10087 )

1. What are the two types of charge called?

2. Provide evidence for the existence of two types of charge.

3. Fill in the blanks: The electrostatic force between like charges is while the electrostatic force

between opposite charges is .

4. I have two positively charged metal balls placed 2 m apart.

a. Is the electrostatic force between the balls attractive or repulsive?

b. If I now move the balls so that they are 1 m apart, what happens to the strength of the electrostatic

force between them?

5. I have 2 charged spheres each hanging from string as shown in the picture below.

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Choose the correct answer from the options below: The spheres will

a. swing towards each other due to the attractive electrostatic force between them.

b. swing away from each other due to the attractive electrostatic force between them.

c. swing towards each other due to the repulsive electrostatic force between them.

d. swing away from each other due to the repulsive electrostatic force between them.

6. Describe how objects (insulators) can be charged by contact or rubbing.

7. You are given a perspex ruler and a piece of cloth.

a. How would you charge the perspex ruler?

b. Explain how the ruler becomes charged in terms of charge.

c. How does the charged ruler attract small pieces of paper?

8. (IEB 2005/11 HG) An uncharged hollow metal sphere is placed on an insulating stand. A positively charged

rod is brought up to touch the hollow metal sphere at P as shown in the diagram below. It is then moved

away from the sphere.

Where is the excess charge distributed on the sphere after the rod has been removed?

a. It is still located at point P where the rod touched the sphere.

b. It is evenly distributed over the outer surface of the hollow sphere.

c. It is evenly distributed over the outer and inner surfaces of the hollow sphere.

d. No charge remains on the hollow sphere.

9. What is the process called where molecules in an uncharged object are caused to align in a particular

direction due to an external charge?

10. Explain how an uncharged object can be attracted to a charged object. You should use diagrams to illustrate

your answer.

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11. Explain how a stream of water can be attracted to a charged rod.

Find the answers with the shortcodes:

(1.) lqs (2.) lqo (3.) lqA (4.) lqG (5.) lqf (6.) lqw

(7.) lqv (8.) lqd (9.) lqp (10.) la2 (11.) lqP

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Electric Circuits

1.174 Electric Circuits

(section shortcode: P10089 )

People all over the world depend on electricity to provide power for most appliances in the home and at work. For

example, fluorescent lights, electric heating and cooking (on electric stoves), all depend on electricity to work. To

realise just how big an impact electricity has on our daily lives, just think about what happens when there is a

power failure or load shedding.

1.174.1 Discussion : Uses of electricity

With a partner, take the following topics and, for each topic, write down at least 5 items/appliances/machines

which need electricity to work. Try not to use the same item more than once.

• At home

• At school

• At the hospital

• In the city

Once you have finished making your lists, compare with the lists of other people in your class. (Save your lists

somewhere safe for later because there will be another activity for which you’ll need them.)

When you start comparing, you should notice that there are many different items which we use in our daily lives

which rely on electricity to work!

TIP: Safety Warning: We believe in experimenting and learning about physics at

every opportunity, BUT playing with electricity and electrical appliances can be EX-

TREMELY DANGEROUS! Do not try to build homemade circuits alone. Make sure

you have someone with you who knows if what you are doing is safe. Normal electrical

outlets are dangerous. Treat electricity with respect in your everyday life. Do not touch

exposed wires and do not approach downed power lines.

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1.174.2 Closed circuits

In the following activity we will investigate what is needed to cause charge to flow in an electric circuit.

Experiment : Closed circuits

Aim: To determine what is required to make electrical charges flow. In this experiment, we will use a lightbulb to

check whether electrical charge is flowing in the circuit or not. If charge is flowing, the lightbulb should glow. On

the other hand, if no charge is flowing, the lightbulb will not glow.

Apparatus: You will need a small lightbulb which is attached to a metal conductor (e.g. a bulb from a school

electrical kit), some connecting wires and a battery.

Method: Take the apparatus items and try to connect them in a way that you cause the light bulb to glow (i.e.

charge flows in the circuit).

Questions:

1. Once you have arranged your circuit elements to make the lightbulb glow, draw your circuit.

2. What can you say about how the battery is connected? (i.e. does it have one or two connecting leads

attached? Where are they attached?)

3. What can you say about how the light bulb is connected in your circuit? (i.e. does it connect to one or two

connecting leads, and where are they attached?)

4. Are there any items in your circuit which are not attached to something? In other words, are there any gaps

in your circuit?

Write down your conclusion about what is needed to make an electric circuit work and charge to flow.

In the experiment above, you will have seen that the light bulb only glows when there is a closed circuit i.e. there

are no gaps in the circuit and all the circuit elements are connected in a closed loop. Therefore, in order for

charges to flow, a closed circuit and an energy source (in this case the battery) are needed. (Note: you do not

have to have a lightbulb in the circuit! We used this as a check that charge was flowing.)

Definition: Electric circuit

An electric circuit is a closed path (with no breaks or gaps) along which electrical charges

(electrons) flow powered by an energy source.

1.174.3 Representing electric circuits

Components of electrical circuits

Some common elements (components) which can be found in electrical circuits include light bulbs, batteries,

connecting leads, switches, resistors, voltmeters and ammeters. You will learn more about these items in later

sections, but it is important to know what their symbols are and how to represent them in circuit diagrams. Below

is a table with the items and their symbols:

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Component Symbol Usage

light bulb glows when charge moves through

it

battery

Figure 23.2: The longer

line shows the positive

(+) side of the battery,

the shorter line shows

the (-) side of the battery.

provides energy for charge to

move - conventional current flow

from positive to negative through a

circuit

switch allows a circuit to be open or

closed

resistor resists the flow of charge

OR

continued on next page

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voltmeter measures potential difference

ammeter measures current in a circuit

connecting lead connects circuit elements together

Table 23.1

Circuit diagrams

Definition: Representing circuits

A physical circuit is the electric circuit you create with real components.

A circuit diagram is a drawing which uses symbols to represent the different components

in the physical circuit.

We use circuit diagrams to represent circuits because they are much simpler and more general than drawing the

physical circuit because they only show the workings of the electrical components. You can see this in the two

pictures below. The first picture shows the physical circuit for an electric torch. You can see the light bulb, the

batteries, the switch and the outside plastic casing of the torch. The picture is actually a cross-section of the torch

so that we can see inside it.

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Figure 23.9: Physical components of an electric torch. The dotted line shows the path of the electrical circuit.

Below is the circuit diagram for the electric torch. Now the light bulb is represented by its symbol, as are the

batteries, the switch and the connecting wires. It is not necessary to show the plastic casing of the torch since

it has nothing to do with the electric workings of the torch. You can see that the circuit diagram is much simpler

than the physical circuit drawing!

Figure 23.10: Circuit diagram of an electric torch.

Series and parallel circuits

There are two ways to connect electrical components in a circuit: in series or in parallel.

Definition: Series circuit

In a series circuit, the charge flowing from the battery can only flow along a single path to

return to the battery.

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Definition: Parallel circuit

In a parallel circuit, the charge flowing from the battery can flow along multiple paths to

return to the battery.

The picture below shows a circuit with two resistors connected in series on the left and a circuit with two resistors

connected in parallel on the right. In the series circiut, the charge path from the battery goes through every

component before returning to the battery. In the parallel circuit, there is more than one path for the charge to

flow from the battery through one of the components and back to the battery.

This simulation allows you to experiment with building circuits. (Simulation: lbK)

Exercise 23.1: Drawing circuits I Draw the circuit diagram for a circuit

which has the following components:

1. 1 battery

2. 1 lightbulb connected in series

3. 2 resistors connected in parallel

Solution to Exercise

Step 1.

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Exercise 23.2: Drawing circuits II Draw the circuit diagram for a circuit

which has the following components:

1. 3 batteries in series

2. 1 lightbulb connected in parallel with 1 resistor

3. a switch in series with the batteries

Solution to Exercise

Step 1.

Circuits

1. Using physical components, set up the physical circuit which is described by the circuit diagram below and

then draw the physical circuit:

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2. Using physical components, set up a closed circuit which has one battery and a light bulb in series with a

resistor.

a. Draw the physical circuit.

b. Draw the resulting circuit diagram.

c. How do you know that you have built a closed circuit? (What happens to the light bulb?)

d. If you add one more resistor to your circuit (also in series), what do you notice? (What happens to the

light from the light bulb?)

e. Draw the new circuit diagram which includes the second resistor.

3. Draw the circuit diagram for the following circuit: 2 batteries and a switch in series, and 1 lightbulb which is

in parallel with two resistors.

a. Now use physical components to set up the circuit.

b. What happens when you close the switch? What does does this mean about the circuit?

c. Draw the physical circuit.

Find the answers with the shortcodes:

(1.) lqZ (2.) lqB (3.) lbY

Discussion : Alternative Energy

At the moment, most electric power is produced by burning fossil fuels such as coal and oil. In South Africa,

our main source of electric power is coal-burning power stations. (We also have one nuclear power plant called

Koeberg in the Western Cape). However, burning fossil fuels releases large amounts of pollution into the earth’s

atmosphere and contributes to global warming. Also, the earth’s fossil fuel reserves (especially oil) are starting

to run low. For these reasons, people all across the world are working to find alternative/other sources of energy

and on ways to conserve/save energy. Other sources of energy include wind power, solar power (from the sun),

and hydro-electric power (from water, e.g. dammed rivers) among others.

With a partner, take out the lists you made earlier of the item/appliances/machines which used electricity in the

following environments. For each item, try to think of an alternative AND a way to conserve or save power.

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For example, if you had a flourescent light as an item used in the home, then:

• Alternative: use candles at suppertime to reduce electricity consumption

• Conservation: turn off lights when not in a room, or during the day.

Topics:

• At home

• At school

• At the hospital

• In the city

Once you have finished making your lists, compare with the lists of other people in your class.

1.175 Potential Difference

(section shortcode: P10090 )

1.175.1 Potential Difference

When a circuit is connected and complete, charge can move through the circuit. Charge will not move unless

there is something to make it move. Think of it as though charge is at rest and something has to push it along.

This means that work needs to be done to make charge move. A force acts on the charges, doing work, to make

them move. The force is provided by the battery in the circuit.

We call the moving charge "current" and we will talk about this later.

The amount of work done to move a charge from one point to another point in a circuit is called the potential

difference between those two points. You can think of it as a difference in the potential energy of the charge

due to its position in the circuit. The difference in potential energy, called potential difference, is equal to the

amount of work done to move the charge between the two points.Just like with gravity, when you raise an object

above the ground it has gravitational potential energy due to its position, the same goes for a charge in a circuit

and electrical potential energy. Potential difference is measured between or across two points. We do not say

potential difference through something.

Definition: Potential Difference

Electrical potential difference is the difference in electrical potential energy per unit charge

between two points. The unit of potential difference is the volta (V). The potential difference

of a battery is the voltage measured across it when current is flowing through it.

anamed after the Italian physicist Alessandro Volta (1745–1827)

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The unit of potential difference is the volt (V), which is the same as 1 joule per coulomb, the amount of work done

per unit charge. Electrical potential difference is also called voltage.

1.175.2 Potential difference and emf

We use an instrument called a voltmeter to measure the potential difference between two points in a circuit. It

must be connected across the two points, in parallel to that portion of the circuit as shown in the diagram below.

Figure 23.16: A voltmeter should be connected in parallel in a circuit.

EMF

When you use a voltmeter to measure the potential difference across (or between) the terminals of a battery,

when no current is flowing through the battery, you are measuring the electromotive force (emf) of the battery.

This is how much potential energy the battery has to make charges move through the circuit. It is a measure of

how much chemical potential energy can be transferred to electrical energy in the battery. This driving potential

energy is equal to the total potential energy drops in the circuit. This means that the voltage across the battery is

equal to the sum of the voltages in the circuit.

Definition: emf

The emf (electromotive force) is the voltage measured across the terminals of a battery

when no current is flowing through the battery.

You have now learnt that the emf is the voltage measured across the terminals of a battery when there is no

current flowing through it and that the potential difference of a battery is the voltage measured across it when

there is current flowing through it. So how do these two quantities compare with each other?

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Experiment : Investigate the emf and potential difference of a battery

Aim: To measure the emf and potential difference across a battery in a circuit

Apparatus: A battery, connecting wires, a light bulb, a switch, a voltmeter.

Method: Set up a circuit with a battery, a lightbulb and a switch connected in series.

First we will measure the emf of the battery when no current is flowing. Make sure that the switch is open and the

lightbulb is not glowing. This is how we check that there is no current flowing through the circuit. Then connect

the voltmeter across the terminals of the battery. Make sure that the voltmeter is set to the largest scale when

you first start measuring. Write down the voltage you measure. This is the emf. Disconnect the voltmeter.

Second we will measure the potential difference across the battery. This time, make sure the switch is closed

and that the lightbulb is glowing. This means that there is current flowing through the circuit. Now connect the

voltmeter again across the terminals of the battery. Again make sure that the voltmeter is set on its largest scale.

Write down the voltage you measure. This is called the potential difference of the battery. How does it compare

to the other value you measured when there was no current flowing?

Summary:You will have noticed that the voltages you measured when there was no current flowing (emf) and

when there was current flowing (potential difference) were different. The emf was a higher value than the potential

difference. Discuss with your classmates and teacher why you think this might happen.

Batteries all have some internal resistance to charges moving through them and therefore some work is required

to move the charges through the battery itself. This internal resistance causes the potential difference across the

battery terminals to be slightly less than the emf.

In the following exercises, we will assume that we have ’perfect’ batteries with no internal resistance. In this

special case, the potential difference of the battery and its emf will be the same. We can use this information to

solve problems in which the voltages across elements in a circuit add up to the emf.

EMF = Vtotal (23.1)

Exercise 23.3: Voltages I What is the voltage across the resistor in the

circuit shown?

Solution to Exercise

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Step 1. We have a circuit with a battery and one resistor. We know the

voltage across the battery. We want to find the voltage across the

resistor.

Vbattery = 2V (23.2)

Step 2. We know that the voltage across the battery must be equal to the

total voltage across all other circuit components.

Vbattery = Vtotal (23.3)

There is only one other circuit component, the resistor.

Vtotal = V1 (23.4)

This means that the voltage across the battery is the same as the

voltage across the resistor.

Vbattery = Vtotal = V1 (23.5)

Vbattery = Vtotal = V1 (23.6)

V1 = 2V (23.7)

Exercise 23.4: Voltages II What is the voltage across the unknown

resistor in the circuit shown?

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Solution to Exercise

Step 1. We have a circuit with a battery and two resistors. We know the

voltage across the battery and one of the resistors. We want to find

that voltage across the resistor.

Vbattery = 2V (23.8)

VA = 1V (23.9)

Step 2. We know that the voltage across the battery must be equal to the

total voltage across all other circuit components that are in series.

Vbattery = Vtotal (23.10)

The total voltage in the circuit is the sum of the voltages across the

individual resistors

Vtotal = VA + VB (23.11)

Using the relationship between the voltage across the battery and

total voltage across the resistors

Vbattery = Vtotal (23.12)

Vbattery = V1 + Vresistor

2V = V1 + 1V

V1 = 1V

(23.13)

Exercise 23.5: Voltages III What is the voltage across the unknown

resistor in the circuit shown?

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Solution to Exercise

Step 1. We have a circuit with a battery and three resistors. We know the

voltage across the battery and two of the resistors. We want to find

that voltage across the unknown resistor.

Vbattery = 7V (23.14)

Vknown = VA + VC

= 1V + 4V(23.15)

Step 2. We know that the voltage across the battery must be equal to the

total voltage across all other circuit components that are in series.

Vbattery = Vtotal (23.16)

The total voltage in the circuit is the sum of the voltages across the

individual resistors

Vtotal = VB + Vknown (23.17)

Using the relationship between the voltage across the battery and

total voltage across the resistors

Vbattery = Vtotal (23.18)

Vbattery = VB + Vknown

7V = VB + 5V

VB = 2V

(23.19)

Exercise 23.6: Voltages IV What is the voltage across the parallel

resistor combination in the circuit shown? Hint: the rest of the circuit is

the same as the previous problem.

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Solution to Exercise

Step 1. The circuit is the same as the previous example and we know that

the voltage difference between two points in a circuit does not de-

pend on what is between them so the answer is the same as above

Vparallel = 2V.

Step 2. We have a circuit with a battery and five resistors (two in series and

three in parallel). We know the voltage across the battery and two

of the resistors. We want to find that voltage across the parallel

resistors, Vparallel.

Vbattery = 7V (23.20)

Vknown = 1V + 4V (23.21)

Step 3. We know that the voltage across the battery must be equal to the

total voltage across all other circuit components.

Vbattery = Vtotal (23.22)

Voltages only add for components in series. The resistors in paral-

lel can be thought of as a single component which is in series with

the other components and then the voltages can be added.

Vtotal = Vparallel + Vknown (23.23)

Using the relationship between the voltage across the battery and

total voltage across the resistors

Vbattery = Vtotal (23.24)

Vbattery = Vparallel + Vknown

7V = V1 + 5V

Vparallel = 2V

(23.25)

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1.176. CURRENT CHAPTER 1. ELECTRIC CIRCUITS

1.176 Current

(section shortcode: P10091 )

1.176.1 Flow of Charge

We have been talking about moving charge. But how much charge is moving, and how fast is it moving? The

concept that represents this information is called current. Current allows us to quantify the movement of charge.

When we talk about current we talk about how much charge moves past a fixed point in circuit in one second.

Think of charges being pushed around the circuit by the battery; there are charges in the wires but unless there

is a battery they won’t move. When one charge moves, the charges next to it also move. They keep their spacing

as if you had a tube of marbles like in this picture.

If you push one marble into the tube then one must come out the other side. If you look at any point in the tube

and push one marble into the tube, one marble will move past the point you are looking at. This is similar to

charges in the wires of a circuit.

If one charge moves then they all move and the same number move at every point in the circuit. This is due to

the conservation of charge.

1.176.2 Current

Now that we’ve thought about moving charges and visualised what is happening we need to get back to quanti-

fying moving charge. We’ve already said that we call moving charge current but we define it precisely as follows:

Definition: Current

Current is the rate of flow of charge, i.e. the rate at which charges move past a fixed

point in a circuit. We use the symbol I to show current and it is measured in amperes (A).

One ampere is one coulomb of charge moving in one second. The relationship between

current, charge and time is given by:

I =Q

∆t(23.26)

When current flows in a circuit we show this on a diagram by adding arrows. The arrows show the direction of

flow in the circuit. By convention we say that charge flows from the positive end (or terminal) of a battery, through

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the circuit, and back to the negative end (or terminal) of the battery. This is shown in the diagram below. We

measure the current with an instrument called an ammeter.

The arrows in this picture show you the direction that charge will flow in the circuit. Note that the arrows point

from the positive end of the battery, through the circuit, towards the negative end of the battery.

NOTE: Benjamin Franklin made a guess about the direction of charge flow when

rubbing smooth wax with rough wool. He thought that the charges flowed from the wax

to the wool (i.e. from positive to negative) which was opposite to the real direction.

Due to this, electrons are said to have a negative charge and so objects which Ben

Franklin called "negative" (meaning a shortage of charge) really have an excess of

electrons. By the time the true direction of electron flow was discovered, the convention

of ´positive¡ and ´negative¡ had already been so well accepted in the scientific world

that no effort was made to change it.

TIP: A battery does not produce the same amount of current no matter what is con-

nected to it. While the voltage produced by a battery is constant, the amount of current

supplied depends on what is in the circuit.

Exercises: Current

Exercise 23.7 Using the relationship between current, charge, and time,

calculate the current in a circuit which has 0, 8 C of charge passing a

point every second.

Solution to Exercise

Step 1. I = Q∆t

Step 2. Given: Q = 0, 8 C∆t = 1 s (23.27)

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Asked for: I

Step 3.

I = 0,8 C1 s

= 0, 8 A(23.28)

Exercise 23.8 How much charge flows per second in a circuit with a

current of 1, 5 A?

Solution to Exercise

Step 1. Asked for: Charge, Q

Given: Current I = 1, 5 A

∆t = 1 s (23.29)

Step 2. I = Q∆t

Step 3.

Q = I ×∆t

= 1, 5 A× 1 s

= 1, 5 C

Exercise 23.9 If 500 × 103 µC of charge flow past a point in a circuit in

1 second, what is the current in the circuit?

Solution to Exercise

Step 1. Asked for: current, I

Given: charge Q = 500× 103 µC∆t = 1 s

Step 2.

Q = 500× 103µC

= 500× 103 × 10−6C

= 0, 5C

Step 3. I = Q∆t

Step 4.

I = Q∆t

= 0,5C1s

= 0, 5A

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Exercise 23.10 I measure the current in a circuit to be 500mA. How

much charge is flowing per second in the circuit?

Solution to Exercise

Step 1. Asked for: Charge, Q

Given: Current I = 500mA,

∆t = 1sStep 2. 500mA = 500× 10−3AStep 3. I = Q

∆t

Step 4.

Q = I ×∆t

= 500× 10−3A× 1s

= 0, 5 C

1.176.3 Measuring voltage and current in circuits

As we have seen in previous sections, an electric circuit is made up of a number of different components such

as batteries, resistors and light bulbs. There are devices to measure the properties of these components. These

devices are called meters.

For example, one may be interested in measuring the amount of current flowing through a circuit using an amme-

ter or measuring the voltage provided by a battery using a voltmeter. In this section we will discuss the practical

usage of voltmeters and ammeters.

Voltmeter

A voltmeter is an instrument for measuring the voltage between two points in an electric circuit. In analogy with

a water circuit, a voltmeter is like a meter designed to measure pressure difference. Since one is interested in

measuring the voltage between two points in a circuit, a voltmeter must be connected in parallel with the portion

of the circuit on which the measurement is made.

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Figure 23.23: A voltmeter should be connected in parallel in a circuit.

Figure 23.23 shows a voltmeter connected in parallel with a battery. The positive lead of the voltmeter must be

connected closest to the positive end of the battery and the negative lead closest to the negative end of the

battery. The voltmeter may also be used to measure the voltage across a resistor or any other component of a

circuit that has a voltage drop or potential difference.

Ammeter

An ammeter is an instrument used to measure the flow of electric current in a circuit. Since one is interested in

measuring the current flowing through the circuit, the ammeter must be connected in series with the measured

circuit component (Figure 23.24). The positive lead from the ammeter must be connected closest to the positive

end of the battery and the negative lead must be connected closest to the negative end of the battery.

Figure 23.24: An ammeter should be connected in series in a circuit.

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Impact of meters on circuits

A good quality meter used correctly will not significantly change the values it is used to measure. This means

that an ammeter has very low resistance so as to not slow down the flow of charge. A voltmeter has a very high

resistance so that it does not add another parallel pathway to the circuit for the charge to flow along.

Investigation : Using meters

If possible, connect meters in circuits to get used to how to use meters to measure electrical quantities. If the

meters have more than one scale, always connect to the largest scale first so that the meter will not be damaged

by having to measure values that exceed its limits.

The table below summarises the use of each measuring instrument that we discussed and the way it should be

connected to a circuit component.

Instrument Measured Quantity Proper Connection

Voltmeter Voltage In Parallel

Ammeter Current In Series

Table 23.2

1.177 Resistance

(section shortcode: P10092 )

The resistance of a circuit element can be thought of as how much it opposes the flow of electric current in the

circuit.

Definition: Resistance

The resistance of a conductor is defined as the potential difference across it divided by the

current flowing though it. We use the symbol R to show resistance and it is measured in

units called Ohms with the symbol Ω.

1 Ohm = 1Volt

Ampere. (23.30)

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1.177.1 What causes resistance?

We have spoken about resistors that reduce the flow of charge in a conductor. On a microscopic level, electrons

moving through the conductor collide with the particles of which the conductor (metal) is made. When they

collide, they transfer kinetic energy. The electrons lose kinetic energy and slow down. This leads to resistance.

The transferred energy causes the resistor to heat up. You can feel this directly if you touch a cellphone charger

when you are charging a cell phone - the charger gets warm because its circuits have some resistors in them!

All conductors have some resistance. For example, a piece of wire has less resistance than a light bulb, but both

have resistance. A lightbulb is a very thin wire surrounded by a glass housing The high resistance of the filament

(small wire) in a lightbulb causes the electrons to transfer a lot of their kinetic energy in the form of heat3. The

heat energy is enough to cause the filament to glow white-hot which produces light. The wires connecting the

lamp to the cell or battery hardly even get warm while conducting the same amount of current. This is because

of their much lower resistance due to their larger cross-section (they are thicker).

An important effect of a resistor is that it converts electrical energy into heat energy. Light is a by-product of the

heat that is produced.

NOTE: There is a special type of conductor, called a superconductor that has no

resistance, but the materials that make up all known superconductors only start super-

conducting at very low temperatures (approximately -170C).

Why do batteries go flat?

A battery stores chemical potential energy. When it is connected in a circuit, a chemical reaction takes place

inside the battery which converts chemical potential energy to electrical energy which powers the charges (elec-

trons) to move through the circuit. All the circuit elements (such as the conducting leads, resistors and lightbulbs)

have some resistance to the flow of charge and convert the electrical energy to heat and, in the case of the

lightbulb, heat and light. Since energy is always conserved, the battery goes flat when all its chemical potential

energy has been converted into other forms of energy.

1.177.2 Resistors in electric circuits

It is important to understand what effect adding resistors to a circuit has on the total resistance of a circuit and

on the current that can flow in the circuit.

Resistors in series

When we add resistors in series to a circuit, we increase the resistance to the flow of current. There is only

one path along which the current can flow and the current is the same at all places in the series circuit. Take a

look at the diagram below: On the left there is a circuit with a single resistor and a battery. No matter where we

measure the current, it is the same in a series circuit. On the right, we have added a second resistor in series to

the circuit. The total resistance of the circuit has increased and you can see from the reading on the ammeter

that the current in the circuit has decreased and is still the same everywhere in the circuit.

3Flourescent lightbulbs do not use thin wires; they use the fact that certain gases glow when a current flows through them. They are much

more efficient (much less resistance) than lightbulbs.

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Potential difference and resistors in series When resistors are in series, one after the other, there is a potential

difference across each resistor. The total potential difference across a set of resistors in series is the sum of the

potential differences across each of the resistors in the set. This is the same as falling a large distance under

gravity or falling that same distance (difference) in many smaller steps. The total distance (difference) is the

same.

Look at the circuits below. If we measured the potential difference between the black dots in all of these circuits

it would be the same; it is just the potential difference across the battery which is the same as the potential

difference across the rest of the circuit. So we now know the total potential difference is the same across one,

two or three resistors. We also know that some work is required to make charge flow through each one. Each

is a step down in potential energy. These steps add up to the total voltage drop which we know is the difference

between the two dots. The sum of the potential differences across each individual resistor is equal to the potential

difference measured across all of them together. For this reason, series circuits are sometimes called voltage

dividers.

Let us look at this in a bit more detail. In the picture below you can see what the different measurements for 3

identical resistors in series could look like. The total voltage across all three resistors is the sum of the voltages

across the individual resistors.

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1.177. RESISTANCE CHAPTER 1. ELECTRIC CIRCUITS

Equivalent Series Resistance

When there is more than one resistor in a circuit, we are usually able to calculate the total combined resitance

of all the resistors. The resistance of the single resistor is known as equivalent resistance or total resistance.

Consider a circuit consisting of three resistors and a single cell connected in series.

We can define the total resistance in a series circuit as:

Definition: Equivalent resistance in a series circuit, Rs

For n resistors in series the equivalent resistance is:

Rs = R1 +R2 +R3 + · · ·+Rn (23.31)

The more resistors we add in series, the higher the equivalent resistance in the circuit. Since the resistors act as

obstacles to the flow of charge through the circuit, the current in the circuit is reduced. Therefore, the higher the

resistance in the circuit, the lower the current through the battery and the circuit. We say that the current in the

battery is inversely proportional to the resistance in the circuit. Let us apply the rule of equivalent resistance in a

series circuit to the following circuit.

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The resistors are in series, therefore:

Rs = R1 +R2 +R3

= 3Ω + 10Ω + 5Ω

= 18Ω

(23.32)

Experiment : Current in Series Circuits

Aim: To determine the effect of multiple resistors on current in a circuit

Apparatus:

• Battery

• Resistors

• Light bulb

• Wires

Method:

1. Construct the following circuits

2. Rank the three circuits in terms of the brightness of the bulb.

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Conclusions: The brightness of the bulb is an indicator of how much current is flowing. If the bulb gets brighter

because of a change then more current is flowing. If the bulb gets dimmer less current is flowing. You will find

that the more resistors you have the dimmer the bulb.

Exercise 23.11: Equivalent series resistance I Two 10 kΩ resistors

are connected in series. Calculate the equivalent resistance.

Solution to Exercise

Step 1. Since the resistors are in series we can use:

Rs = R1 +R2 (23.33)

Step 2.

Rs = R1 +R2

= 10kΩ + 10kΩ

= 20kΩ

(23.34)

Step 3. The equivalent resistance of two 10 kΩ resistors connected in se-

ries is 20 kΩ.

Exercise 23.12: Equivalent series resistance II Two resistors are

connected in series. The equivalent resistance is 100 Ω. If one resistor

is 10 Ω, calculate the value of the second resistor.

Solution to Exercise

Step 1. Since the resistors are in series we can use:

Rs = R1 +R2 (23.35)

We are given the value of Rs and R1.

Step 2.

Rs = R1 +R2

∴ R2 = Rs −R1

= 100Ω− 10Ω

= 90Ω

(23.36)

Step 3. The second resistor has a resistance of 90 Ω.

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Khan academy video on circuits - 2 (Video: P10093 )

Resistors in parallel

In contrast to the series case, when we add resistors in parallel, we create more paths along which current can

flow. By doing this we decrease the total resistance of the circuit!

Take a look at the diagram below. On the left we have the same circuit as shown on the left in Figure 23.25 with a

battery and a resistor. The ammeter shows a current of 1 ampere. On the right we have added a second resistor

in parallel to the first resistor. This has increased the number of paths (branches) the charge can take through

the circuit - the total resistance has decreased. You can see that the current in the circuit has increased. Also

notice that the current in the different branches can be different (in this case 1 A and 2 A) but must add up to the

current through the battery (3 A). Since the total current in the circuit is equal to the sum of the currents in the

parallel branches, a parallel circuit is sometimes called a current divider.

Potential difference and parallel resistorsWhen resistors are connected in parallel the start and end points

for all the resistors are the same. These points have the same potential energy and so the potential difference

between them is the same no matter what is put in between them. You can have one, two or many resistors

between the two points, the potential difference will not change. You can ignore whatever components are

between two points in a circuit when calculating the difference between the two points.

Look at the following circuit diagrams. The battery is the same in all cases. All that changes is that more resistors

are added between the points marked by the black dots. If we were to measure the potential difference between

the two dots in these circuits we would get the same answer for all three cases.

Let’s look at two resistors in parallel more closely. When you construct a circuit you use wires and you might think

that measuring the voltage in different places on the wires will make a difference. This is not true. The potential

difference or voltage measurement will only be different if you measure a different set of components. All points

on the wires that have no circuit components between them will give you the same measurements.

All three of the measurements shown in the picture below (i.e. A–B, C–D and E–F) will give you the same voltage.

The different measurement points on the left (i.e. A, E, C) have no components between them so there is no

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1.177. RESISTANCE CHAPTER 1. ELECTRIC CIRCUITS

change in potential energy. Exactly the same applies to the different points on the right (i.e. B, F, D). When you

measure the potential difference between the points on the left and right you will get the same answer.

Definition: Equivalent resistance of two parallel resistor, Rp

For 2 resistors in parallel with resistances R1 and R2, the equivalent resistance is:

Rp =R1R2

R1 +R2(23.37)

Equivalent parallel resistance

Consider a circuit consisting of a single cell and three resistors that are connected in parallel.

Using what we know about voltage and current in parallel circuits we can define the equivalent resistance of

several resistors in parallel as:

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Definition: Equivalent resistance in a parallel circuit, Rp

For n resistors in parallel, the equivalent resistance is:

1

Rp

=

(

1

R1+

1

R2+

1

R3+ · · ·+ 1

Rn

)

(23.38)

Let us apply this formula to the following circuit.

What is the total resistance in the circuit?

1Rp

=(

1R1

+ 1R2

+ 1R3

)

=(

110 Ω + 1

2 Ω + 11 Ω

)

=(

1+5+1010

)

=(

1610

)

∴ Rp = 0, 625Ω

(23.39)

Experiment : Current in Parallel Circuits

Aim: To determine the effect of multiple resistors on current in a circuit

Apparatus:

• Battery

• Resistors

• Light bulb

• Wires

Method:

1. Construct the following circuits

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1.177. RESISTANCE CHAPTER 1. ELECTRIC CIRCUITS

2. Rank the three circuits in terms of the brightness of the bulb.

Conclusions: The brightness of the bulb is an indicator of how much current is flowing. If the bulb gets brighter

because of a change then more current is flowing. If the bulb gets dimmer less current is flowing. You will find

that the more resistors you have the brighter the bulb.

Why is this the case? Why do more resistors make it easier for charge to flow in the circuit? It is because they

are in parallel so there are more paths for charge to take to move. You can think of it like a highway with more

lanes, or the tube of marbles splitting into multiple parallel tubes. The more branches there are, the easier it is for

charge to flow. You will learn more about the total resistance of parallel resistors later but always remember that

more resistors in parallel mean more pathways. In series the pathways come one after the other so it does not

make it easier for charge to flow.

Exercise 23.13 Two 8 kΩ resistors are connected in parallel. Calculate

the equivalent resistance.

Solution to Exercise

Step 1. Since the resistors are in parallel we can use:

1

Rp

=1

R1+

1

R2(23.40)

Step 2.1Rp

= 1R1

+ 1R2

= 18 k Ω + 1

10 k Ω

Rp = 28

= 4kΩ

(23.41)

Step 3. The equivalent resistance of two 8 kΩ resistors connected in paral-

lel is 4 kΩ.

Exercise 23.14 Two resistors are connected in parallel. The equivalent

resistance is 100 Ω. If one resistor is 150 Ω, calculate the value of the

second resistor.

Solution to Exercise

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CHAPTER 1. ELECTRIC CIRCUITS 1.178.

Step 1. Since the resistors are in parallel we can use:

1

Rp

=1

R1+

1

R2(23.42)

We are given the value of Rp and R1.

Step 2.1Rp

= 1R1

+ 1R2

∴1R2

= 1Rp

− 1R1

= 1100 Ω − 1

150 Ω

= 3−2300

= 1300

R2 = 300Ω

(23.43)

Step 3. The second resistor has a resistance of 300 Ω.

Khan academy video on circuits - 3 (Video: P10094 )

Resistance

1. What is the unit of resistance called and what is its symbol?

2. Explain what happens to the total resistance of a circuit when resistors are added in series?

3. Explain what happens to the total resistance of a circuit when resistors are added in parallel?

4. Why do batteries go flat?

Find the answers with the shortcodes:

(1.) lqk (2.) lq0 (3.) lq8 (4.) lq9

1.178

(section shortcode: P10095 )

Khan academy video on circuits - 4 (Video: P10096 )

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1.179. EXERCISES - ELECTRIC CIRCUITS CHAPTER 1. ELECTRIC CIRCUITS

The following presentation summarizes the concepts covered in this chapter.

(Presentation: P10097 )

1.179 Exercises - Electric circuits

(section shortcode: P10098 )

1. Write definitions for each of the following:

a. resistor

b. coulomb

c. voltmeter

2. Draw a circuit diagram which consists of the following components:

a. 2 batteries in parallel

b. an open switch

c. 2 resistors in parallel

d. an ammeter measuring total current

e. a voltmeter measuring potential difference across one of the parallel resistors

3. Complete the table below:

Quantity Symbol Unit of meaurement Symbol of unit

e.g. Distance e.g. d e.g. kilometer e.g. km

Resistance

Current

Potential difference

Table 23.3

4. Draw a diagram of a circuit which contains a battery connected to a lightbulb and a resistor all in series.

a. Also include in the diagram where you would place an ammeter if you wanted to measure the current

through the lightbulb.

b. Draw where and how you would place a voltmeter in the circuit to measure the potential difference

across the resistor.

5. Thandi wants to measure the current through the resistor in the circuit shown below and sets up the circuit

as shown below. What is wrong with her circuit setup?

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6. (SC 2003/11) The emf of a battery can best be explained as the · · ·a. rate of energy delivered per unit current

b. rate at which charge is delivered

c. rate at which energy is delivered

d. charge per unit of energy delivered by the battery

7. (IEB 2002/11 HG1) Which of the following is the correct definition of the emf of a battery?

a. It is the product of current and the external resistance of the circuit.

b. It is a measure of the cell’s ability to conduct an electric current.

c. It is equal to the “lost volts” in the internal resistance of the circuit.

d. It is the power supplied by the battery per unit current passing through the battery.

8. (IEB 2005/11 HG) Three identical light bulbs A, B and C are connected in an electric circuit as shown in the

diagram below.

a. How bright is bulb A compared to B and C?

b. How bright are the bulbs after switch S has been opened?

c. How do the currents in bulbs A and B change when switch S is opened?

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1.179. EXERCISES - ELECTRIC CIRCUITS CHAPTER 1. ELECTRIC CIRCUITS

Current in A Current in B

(a) decreases increases

(b) decreases decreases

(c) increases increases

(d) increases decreases

Table 23.4

9. (IEB 2004/11 HG1) When a current I is maintained in a conductor for a time of t, how many electrons with

charge e pass any cross-section of the conductor per second?

a. It

b. It/e

c. Ite

d. e/It

Find the answers with the shortcodes:

(1.) lqX (2.) lqI (3.) lq5 (4.) lTw (5.) lTv (6.) lqn

(7.) lqR (8.) lqN (9.) lqQ

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Safety

.1 Introduction

(section shortcode: S10000 )

A laboratory (be it for physics, chemistry or other sciences) can be a very dangerous and daunting place. How-

ever, if you follow a few simple guidelines you can safely carry out experiments in the lab without endangering

yourself or others around you.

.2 General safety rules

(section shortcode: S10001 )

The following are some of the general guidelines and rules that you should always observe when working in a

laboratory.

1. Do not eat or drink in the lab. Do not use lab glassware to eat or drink from.

2. Always behave responsibly in the lab. Do not run around or play practical jokes.

3. In case of accidents or chemical spills call your teacher at once.

4. Always check with your teacher how to dispose of waste. Chemicals should not be disposed of down the

sink.

5. Only perform the experiments that your teacher instructs you to. Never mix chemicals for fun.

6. Never perform experiments alone.

7. Always check the safety data of any chemicals you are going to use.

8. Follow the given instructions exactly. Do not mix up steps or try things in a different order.

9. Be alert and careful when handling chemicals, hot glassware, etc.

10. Ensure all bunsen burners are turned off at the end of the practical and all chemical containers are sealed.

.3 Hazard signs

(section shortcode: S10002 )

The image below lists some of the common hazards signs that you may encounter. You should know what all of

these mean.

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.4. NOTES AND INFORMATION APPENDIX . SAFETY

.4 Notes and information

(section shortcode: S10003 )

You can find safety data sheets at Merck4 . You should always look at these data sheets anytime you work with

a new chemical.

You should always try dispose of chemicals correctly and safely. Many chemicals cannot simply be washed down

the sink.

4http://www.merck-chemicals.co.za/safety-data-sheets/c_O_Sb.s1LQz0AAAEWVOYfVhTo

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Glossary

A Acceleration

Acceleration is the rate of change of velocity.

Acid rain

Acid rain refers to the deposition of acidic components in rain, snow and dew. Acid rain occurs when

sulphur dioxide and nitrogen oxides are emitted into the atmosphere, undergo chemical

transformations and are absorbed by water droplets in clouds. The droplets then fall to earth as rain,

snow, mist, dry dust, hail, or sleet. This increases the acidity of the soil and affects the chemical

balance of lakes and streams.

Amplitude

The amplitude is the maximum displacement from equilibrium. For a longitudinal wave which is a

pressure wave this would be the maximum increase (or decrease) in pressure from the equilibrium

pressure that is cause when a peak (or trough) passes a point.

Amplitude

The amplitude is the maximum displacement of a particle from its equilibrium position.

Amplitude

The amplitude of a pulse is a measurement of how far the medium is displaced from rest.

Anti-Node

An anti-node is a point on standing a wave where maximum displacement takes place. A free end of a

rope is an anti-node.

Atomic mass number (A)

The number of protons and neutrons in the nucleus of an atom

Atomic number (Z)

The number of protons in an atom

Atomic orbital

An atomic orbital is the region in which an electron may be found around a single atom.

Attraction and Repulsion

Like poles of magnets repel each other whilst unlike poles attract each other.

Average velocity

Average velocity is the total displacement of a body over a time interval.

Avogadro’s number

The number of particles in a mole, equal to 6, 022× 1023. It is also sometimes referred to as the

number of atoms in 12 g of carbon-12.

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.4. NOTES AND INFORMATION APPENDIX . SAFETY

B Boiling point

The temperature at which a liquid changes its phase to become a gas. The process is called

evaporation and the reverse process is called condensation

C Chemical bond

A chemical bond is the physical process that causes atoms and molecules to be attracted to each

other, and held together in more stable chemical compounds.

Chemical change

The formation of new substances in a chemical reaction. One type of matter is changed into

something different.

Compound

A compound is a group of two or more atoms that are attracted to each other by relatively strong

forces or bonds.

Compound

A substance made up of two or more elements that are joined together in a fixed ratio.

Compression

A compression is a region in a longitudinal wave where the particles are closest together.

Concentration

Concentration is a measure of the amount of solute that is dissolved in a given volume of liquid. It is

measured in mol · dm−3. Another term that is used for concentration is molarity (M)

Conductivity

Conductivity is a measure of a solution’s ability to conduct an electric current.

Conductors and insulators

A conductor allows the easy movement or flow of something such as heat or electrical charge through

it. Insulators are the opposite to conductors because they inhibit or reduce the flow of heat, electrical

charge, sound etc through them.

Conservation of energy principle

Energy cannot be created or destroyed. It can only be changed from one form to another.

Conservation of Energy

The Law of Conservation of Energy: Energy cannot be created or destroyed, but is merely changed

from one form into another.

Conservation of Mechanical Energy

Law of Conservation of Mechanical Energy: The total amount of mechanical energy in a closed

system remains constant.

Constructive interference

Constructive interference is when two pulses meet, resulting in a bigger pulse.

Core electrons

All the electrons in an atom, excluding the valence electrons

Covalent bond

Covalent bonding is a form of chemical bonding where pairs of electrons are shared between atoms.

Current

Current is the rate of flow of charge, i.e. the rate at which charges move past a fixed point in a circuit.

We use the symbol I to show current and it is measured in amperes (A). One ampere is one coulomb

of charge moving in one second. The relationship between current, charge and time is given by:

I =Q

∆t(23.26)

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D Density

Density is a measure of the mass of a substance per unit volume.

Destructive interference

Destructive interference is when two pulses meet, resulting in a smaller pulse.

Displacement

Displacement is the change in an object’s position.

Dissociation

Dissociation in chemistry and biochemistry is a general process in which ionic compounds separate or

split into smaller molecules or ions, usually in a reversible manner.

E Electric circuit

An electric circuit is a closed path (with no breaks or gaps) along which electrical charges (electrons)

flow powered by an energy source.

Electrolyte

An electrolyte is a substance that contains free ions and behaves as an electrically conductive

medium. Because they generally consist of ions in solution, electrolytes are also known as ionic

solutions.

Electron configuration

Electron configuration is the arrangement of electrons in an atom, molecule or other physical structure.

Element

An element is a substance that cannot be broken down into other substances through chemical

means.

emf

The emf (electromotive force) is the voltage measured across the terminals of a battery when no

current is flowing through the battery.

Empirical formula

The empirical formula of a chemical compound gives the relative number of each type of atom in that

compound.

Empirical formula

This is a way of expressing the relative number of each type of atom in a chemical compound. In most

cases, the empirical formula does not show the exact number of atoms, but rather the simplest ratio of

the atoms in the compound.

Equivalent resistance in a parallel circuit, Rp

For n resistors in parallel, the equivalent resistance is:

1

Rp

=

(

1

R1+

1

R2+

1

R3+ · · ·+ 1

Rn

)

(23.38)

Equivalent resistance in a series circuit, Rs

For n resistors in series the equivalent resistance is:

Rs = R1 +R2 +R3 + · · ·+Rn (23.31)

Equivalent resistance of two parallel resistor, Rp

For 2 resistors in parallel with resistances R1 and R2, the equivalent resistance is:

Rp =R1R2

R1 +R2(23.37)

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F Frame of Reference

A frame of reference is a reference point combined with a set of directions.

Frequency

The frequency is the number of successive peaks (or troughs) passing a given point in 1 second.

Frequency

The frequency of a wave is the number of wavelengths per second.

G Gradient

The gradient of a line can be calculated by dividing the change in the y-value by the change in the

x-value.

m = ∆y∆x

H Heterogeneous mixture

A heterogeneous mixture is one that is non-uniform and the different components of the mixture can

be seen.

Homogeneous mixture

A homogeneous mixture is one that is uniform, and where the different components of the mixture

cannot be seen.

I Instantaneous velocity

Instantaneous velocity is the velocity of a body at a specific instant in time.

Intermolecular force

A force between molecules, which holds them together.

Intramolecular force

The force between the atoms of a molecule, which holds them together.

Ion

An ion is a charged atom. A positively charged ion is called a cation e.g. Na+, and a negatively

charged ion is called an anion e.g. F−. The charge on an ion depends on the number of electrons

that have been lost or gained.

Ion exchange reaction

A type of reaction where the positive ions exchange their respective negative ions due to a driving

force.

Ionic bond

An ionic bond is a type of chemical bond based on the electrostatic forces between two

oppositely-charged ions. When ionic bonds form, a metal donates one or more electrons, due to

having a low electronegativity, to form a positive ion or cation. The non-metal atom has a high

electronegativity, and therefore readily gains electrons to form a negative ion or anion. The two ions

are then attracted to each other by electrostatic forces.

Isotope

The isotope of a particular element is made up of atoms which have the same number of protons as

the atoms in the original element, but a different number of neutrons.

K Kinetic Energy

Kinetic energy is the energy an object has due to its motion.

L Longitudinal waves

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A longitudinal wave is a wave where the particles in the medium move parallel to the direction of

propagation of the wave.

M Magnetism

Magnetism is one of the phenomena by which materials exert attractive or repulsive forces on other

materials.

Medium

A medium is the substance or material in which a wave will move.

Melting point

The temperature at which a solid changes its phase or state to become a liquid. The process is called

melting and the reverse process (change in phase from liquid to solid) is called freezing.

Metallic bond

Metallic bonding is the electrostatic attraction between the positively charged atomic nuclei of metal

atoms and the delocalised electrons in the metal.

Mixture

A mixture is a combination of two or more substances, where these substances are not bonded (or

joined) to each other.

Model

A model is a representation of a system in the real world. Models help us to understand systems and

their properties. For example, an atomic model represents what the structure of an atom could look

like, based on what we know about how atoms behave. It is not necessarily a true picture of the exact

structure of an atom.

Molar mass

Molar mass (M) is the mass of 1 mole of a chemical substance. The unit for molar mass is grams per

mole or g ·mol−1.

Mole

The mole (abbreviation ’n’) is the SI (Standard International) unit for ’amount of substance’. It is

defined as an amount of substance that contains the same number of particles (atoms, molecules or

other particle units) as there are atoms in 12 g carbon.

Molecular formula

The molecular formula of a chemical compound gives the exact number of atoms of each element in

one molecule of that compound.

Molecular formula

This is a concise way of expressing information about the atoms that make up a particular chemical

compound. The molecular formula gives the exact number of each type of atom in the molecule.

N Node

A node is a point on a standing wave where no displacement takes place at any time. A fixed end of a

rope is a node.

P Parallel circuit

In a parallel circuit, the charge flowing from the battery can flow along multiple paths to return to the

battery.

Peaks and troughs

A peak is a point on the wave where the displacement of the medium is at a maximum. A point on the

wave is a trough if the displacement of the medium at that point is at a minimum.

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Period (T)

The period (T) is the time taken for two successive peaks (or troughs) to pass a fixed point.

Period

The period of a wave is the time taken by the wave to move one wavelength.

Photon

A photon is a quantum (energy packet) of light.

Physical change

A change that can be seen or felt, but that doesn’t involve the break up of the particles in the reaction.

During a physical change, the form of matter may change, but not its identity. A change in temperature

is an example of a physical change.

Physical Quantity

A physical quantity is anything that you can measure. For example, length, temperature, distance and

time are physical quantities.

Planck’s constant

Planck’s constant is a physical constant named after Max Planck.

h = 6, 626× 10−34 J · s

Position

Position is a measurement of a location, with reference to an origin.

Potential Difference

Electrical potential difference is the difference in electrical potential energy per unit charge between

two points. The unit of potential difference is the volt5 (V). The potential difference of a battery is the

voltage measured across it when current is flowing through it.

Potential energy

Potential energy is the energy an object has due to its position or state.

Precipitate

A precipitate is the solid that forms in a solution during a chemical reaction.

Prefix

A prefix is a group of letters that are placed in front of a word. The effect of the prefix is to change

meaning of the word. For example, the prefix un is often added to a word to mean not, as in

unnecessary which means not necessary.

Pulse

A pulse is a single disturbance that moves through a medium.

Pulse Speed

Pulse speed is the distance a pulse travels per unit time.

R Rarefaction

A rarefaction is a region in a longitudinal wave where the particles are furthest apart.

Relative atomic mass

Relative atomic mass is the average mass of one atom of all the naturally occurring isotopes of a

particular chemical element, expressed in atomic mass units.

Representing circuits

A physical circuit is the electric circuit you create with real components.

A circuit diagram is a drawing which uses symbols to represent the different components in the

physical circuit.

5named after the Italian physicist Alessandro Volta (1745–1827)

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Resistance

The resistance of a conductor is defined as the potential difference across it divided by the current

flowing though it. We use the symbol R to show resistance and it is measured in units called Ohms

with the symbol Ω.

1 Ohm = 1Volt

Ampere. (23.30)

Resultant of Vectors

The resultant of a number of vectors is the single vector whose effect is the same as the individual

vectors acting together.

S Scalar

A scalar is a quantity that has only magnitude (size).

Series circuit

In a series circuit, the charge flowing from the battery can only flow along a single path to return to the

battery.

SI Units

The name SI units comes from the French Système International d’Unités, which means international

system of units.

Speed of sound

The speed of sound in air, at sea level, at a temperature of 21C and under normal atmospheric

conditions, is 344 m · s−1.

T The law of conservation of mass

The mass of a closed system of substances will remain constant, regardless of the processes acting

inside the system. Matter can change form, but cannot be created or destroyed. For any chemical

process in a closed system, the mass of the reactants must equal the mass of the products.

Transverse Pulse

A pulse where all of the particles disturbed by the pulse move perpendicular (at a right angle) to the

direction in which the pulse is moving.

Transverse wave

A transverse wave is a wave where the movement of the particles of the medium is perpendicular (at a

right angle) to the direction of propagation of the wave.

V Valence electrons

The electrons in the outer energy level of an atom

Valency

The number of electrons in the outer shell of an atom which are able to be used to form bonds with

other atoms.

Vectors

A vector is a quantity that has both magnitude and direction.

Vectors and Scalars

A vector is a physical quantity with magnitude (size) and direction. A scalar is a physical quantity with

magnitude (size) only.

Velocity

Velocity is the rate of change of displacement.

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Viscosity

Viscosity is a measure of how resistant a liquid is to flowing (in other words, how easy it is to pour the

liquid from one container to another).

W Water hardness

Water hardness is a measure of the mineral content of water. Minerals are substances such as calcite,

quartz and mica that occur naturally as a result of geological processes.

Wave

A wave is a periodic, continuous disturbance that consists of a train of pulses.

Wavelength of wave

The wavelength of a wave is the distance between any two adjacent points that are in phase.

Wavelength

The wavelength in a longitudinal wave is the distance between two consecutive points that are in

phase.

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Index of Keywords and Terms

Keywords are listed by the section with that keyword (page numbers are in parentheses). Keywords do not

necessarily appear in the text of the page. They are merely associated with that section. Ex. apples, § 1.1 (1)

Terms are referenced by the page they appear on. Ex. apples, 1

Chemistry

A Acid rain, 155

aqueous, § 10(167)

Atom, § 4(79)

Atomic combinations, § 6(112)

Atomic mass number (A), 61

Atomic number (Z), 60

Atomic orbital, 71

Avogadro’s number, 172

B Boiling point, 46

C change, § 8(148)

Chemical, § 9(148)

Chemical bond, 93

Chemical change, 125

Chemistry, § 2(35), § 3(50), § 4(79), § 6(112),

§ 7(120), § 8(134), § 10(167), § 11(192),

§ 12(203)

Classification, § 2(35)

Compound, 21, 115

Concentration, 188

Conductivity, 157

Conductors and insulators, 29

Conservation of energy principle, 130

Core electrons, 76

Covalent bond, 94

D Density, 47

Dissociation, 152

E Electrolyte, 157

Electron configuration, 73

Element, 21

Empirical formula, 117, 181

F FHSST, § 6(120), § 8(134)

G Gr10, § 2(50)

Grade 10, § 2(35), § 4(79), § 7(120), § 8(134),

§ 9(148), § 10(167), § 12(203)

Grade 11, § 6(112), § 11(192)

H Heterogeneous mixture, 19

Homogeneous mixture, 19

Hydrosphere, § 12(203)

I Intermolecular force, 44

Intramolecular force, 44

Ion, 87

Ion exchange reaction, 160

Ionic bond, 102

Isotope, 64

M Magnetism, 32

Matter, § 2(35), § 3(50)

Melting point, 46

Metallic bond, 106

Mixture, 18

Model, 54

Molar mass, 174

Mole, 172

Molecular formula, 116, 181

Moleculuar, § 3(50)

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P Physical change, 123

Precipitate, 161

Q Quantitative aspects of chemical change,

§ 11(192)

R Reactions, § 10(167)

Relative atomic mass, 67

S Science, § 2(50)

solution, § 10(167)

South Africa, § 2(35), § 3(50), § 4(79), § 6(112),

§ 7(120), § 8(134), § 9(148), § 10(167),

§ 11(192), § 12(203)

T The law of conservation of mass, 139

Theory, § 3(50)

V Valence electrons, 76

Valency, 97

Viscosity, 47

W Water hardness, 154

Physics

A Acceleration, 277

Amplitude, 312, 337, 368

Anti-Node, 357

Attraction and Repulsion, 406

Average velocity, 269

C Compression, 367

Conservation of Energy, 252

Conservation of Mechanical Energy, 252

Constructive interference, 315

Current, 444

D Destructive interference, 316

Displacement, 265

E Electric circuits, § 23(429)

Electromagnetic Radiation, § 20(391)

Electrostatics, § 22(415)

F FHSST, § 15(261)

Frame of Reference, 262

Frequency, 342, 369

G Grade 10, § 13(207), § 14(245), § 15(261),

§ 16(311), § 17(335), § 21(403), § 22(415),

§ 23(429)

Grade 11, § 18(365), § 19(377)

Grade 12, § 20(391)

Gradient, 280

I Instantaneous velocity, 268

K Kinetic Energy, 248

L Longitudinal Waves, § 18(365), 365

M Magnetism, § 21(403)

Mathematics, § 13(207)

Mechanical Energy, § 14(245)

Medium, 311

Motion in One Dimension, § 15(261)

N Node, 357

P Parallel circuit, 433

Peaks and troughs, 337

Period, 369

Period (T), 342

Photon, 391

Physical Science, § 13(207), § 19(377),

§ 21(403), § 22(415), § 23(429)

Physics, § 13(207), § 14(245), § 16(311),

§ 17(335), § 18(365), § 19(377), § 20(391),

§ 23(429)

Planck’s constant, 396

Position, 263

Potential Difference, 437

Potential energy, 245

Pulse, 312

Pulse Speed, 314

R Rarefaction, 367

Representing circuits, 432

Resistance, 449

Resultant of Vectors, 214

S Scalar, 207

Series circuit, 433

Sound, § 19(377)

South Africa, § 13(207), § 14(245), § 15(261),

§ 16(311), § 17(335), § 18(365), § 19(377),

§ 20(391), § 21(403), § 22(415), § 23(429)

Speed of sound, 380

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T Transverse Pulse, 312

transverse pulses, § 16(311)

Transverse wave, 335

Transverse Waves, § 17(335)

V Vectors, § 13(207), 208

Vectors and Scalars, 267

Velocity, 268

W Wave, 335

Wavelength, 368

Wavelength of wave, 341

waves, § 16(311)

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Attributions

Chemistry

Collection: Chemistry Grade 10 [CAPS]

Edited by: Free High School Science Texts Project

URL: http://http://cnx.org/content/col11303/1.4/

License: http://creativecommons.org/licenses/by/3.0/

Module: "Classification of Matter - Grade 10 [CAPS]"

Used here as: "Classification of Matter"

By: Free High School Science Texts Project, Heather Williams

URL: http://http://cnx.org/content/m38118/1.5/

Pages: 35-37

Copyright: Free High School Science Texts Project, Heather Williams

License: http://creativecommons.org/licenses/by/3.0/

Based on: Classification of Matter - Grade 10

By: Rory Adams, Mark Horner, Sarah Blyth, Heather Williams, Free High School Science Texts Project

URL: http://cnx.org/content/m35747/1.6/

Module: "States of Matter and the Kinetic Molecular Theory - Gr10 [CAPS]"

Used here as: "States of matter and the kinetic molecular theory"

By: Free High School Science Texts Project

URL: http://http://cnx.org/content/m38210/1.7/

Pages: 50-51

Copyright: Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Module: "The Atom - Grade 10 [CAPS]"

Used here as: "The Atom"

By: Free High School Science Texts Project

URL: http://http://cnx.org/content/m38126/1.5/

Pages: 79-81

Copyright: Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: The Atom

By: Rory Adams, Free High School Science Texts Project, Heather Williams

URL: http://cnx.org/content/m35750/1.4/

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Module: "The Periodic Table - Grade 10 [CAPS]"

Used here as: "The Periodic Table"

By: Free High School Science Texts Project

URL: http://http://cnx.org/content/m38133/1.6/

Pages: 91-92

Copyright: Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Module: "Chemical Bonding - Grade 10 (11) [CAPS]"

Used here as: "Chemical Bonding"

By: Free High School Science Texts Project

URL: http://http://cnx.org/content/m38131/1.6/

Pages: 112-114

Copyright: Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Atomic Combinations - Grade 11

By: Rory Adams, Free High School Science Texts Project, Heather Williams

URL: http://cnx.org/content/m35862/1.1/

Module: "What are the objects around us made of - Grade 10 [CAPS]"

Used here as: "What are the objects around us made of"

By: Free High School Science Texts Project, Heather Williams

URL: http://http://cnx.org/content/m38120/1.5/

Pages: 120-121

Copyright: Free High School Science Texts Project, Heather Williams

License: http://creativecommons.org/licenses/by/3.0/

Based on: What are the objects around us made of - Grade 10

By: Rory Adams, Free High School Science Texts Project, Heather Williams

URL: http://cnx.org/content/m35959/1.3/

Module: "Physical and Chemical Change - Grade 10 [CAPS]"

Used here as: "Physical and Chemical Change"

By: Free High School Science Texts Project

URL: http://http://cnx.org/content/m38141/1.8/

Pages: 134-135

Copyright: Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Physical and Chemical Change - Grade 10

By: Rory Adams, Free High School Science Texts Project, Heather Williams

URL: http://cnx.org/content/m36249/1.3/

Module: "Representing Chemical Change - Grade 10 [CAPS]"

Used here as: "Representing Chemical Change"

By: Free High School Science Texts Project

URL: http://http://cnx.org/content/m38139/1.5/

Pages: 148-149

Copyright: Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Representing Chemical Change - Grade 10

By: Rory Adams, Heather Williams, Free High School Science Texts Project

URL: http://cnx.org/content/m35749/1.4/

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Module: "Reactions in aqueous solutions - Grade 10 [CAPS]"

Used here as: "Reactions in Aqueous Solutions"

By: Free High School Science Texts Project

URL: http://http://cnx.org/content/m38136/1.6/

Pages: 167-169

Copyright: Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Module: "Quantitative Aspects of Chemical Change - Grade 10 (11) [CAPS]"

Used here as: "Quantitative Aspects of Chemical Change"

By: Free High School Science Texts Project

URL: http://http://cnx.org/content/m38155/1.4/

Pages: 192-195

Copyright: Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Quantitative Aspects of Chemical Change - Grade 11

By: Rory Adams, Free High School Science Texts Project, Heather Williams

URL: http://cnx.org/content/m35939/1.1/

Module: "The Hydrosphere - Grade 10 [CAPS]"

Used here as: "The Hydrosphere"

By: Free High School Science Texts Project

URL: http://http://cnx.org/content/m38138/1.4/

Page: 203

Copyright: Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: The Hydrosphere

By: Rory Adams, Mark Horner, Free High School Science Texts Project, Heather Williams

URL: http://cnx.org/content/m31323/1.7/

Physics

Collection: Physics - Grade 10 [CAPS 2011]

Edited by: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/col11298/1.2/

License: http://creativecommons.org/licenses/by/3.0/

Module: "Vectors - Grade 10 (11) [CAPS]"

Used here as: "Vectors"

By: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/m37829/1.3/

Pages: 207-244

Copyright: Mark Horner, Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Vectors - Grade 11

By: Rory Adams, Free High School Science Texts Project, Mark Horner, Heather Williams

URL: http://cnx.org/content/m32833/1.3/

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Module: "Mechanical Energy - Grade 10 [CAPS]"

Used here as: "Gravity and Mechanical Energy"

By: Free High School Science Texts Project

URL: http://http://cnx.org/content/m37174/1.4/

Pages: 245-260

Copyright: Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Gravity and Mechanical Energy

By: Rory Adams, Free High School Science Texts Project, Sarah Blyth, Heather Williams

URL: http://cnx.org/content/m31482/1.5/

Module: "Motion in One Dimension - Grade 10 [CAPS]"

Used here as: "Motion in One Dimension"

By: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/m37923/1.3/

Pages: 261-309

Copyright: Mark Horner, Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Motion in One Dimension - Grade 10

By: Rory Adams, Free High School Science Texts Project, Wendy Williams, Heather Williams

URL: http://cnx.org/content/m30867/1.4/

Module: "Transverse Pulses - Grade 10 [CAPS]"

Used here as: "Transverse Pulses"

By: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/m37832/1.4/

Pages: 311-334

Copyright: Mark Horner, Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Transverse Pulses - Grade 10

By: Rory Adams, Free High School Science Texts Project, Heather Williams

URL: http://cnx.org/content/m35714/1.4/

Module: "Transverse Waves - Grade 10 [CAPS]"

Used here as: "Transverse Waves"

By: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/m37834/1.5/

Pages: 335-364

Copyright: Mark Horner, Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Transverse Waves - Grade 10

By: Rory Adams, Free High School Science Texts Project, Mark Horner, Heather Williams

URL: http://cnx.org/content/m32635/1.3/

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Module: "Longitudinal Waves - Grade 10 (11) [CAPS]"

Used here as: "Longitudinal Waves"

By: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/m37831/1.5/

Pages: 365-375

Copyright: Mark Horner, Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Longitudinal Waves - Grade 11

By: Rory Adams, Free High School Science Texts Project, Heather Williams

URL: http://cnx.org/content/m30841/1.2/

Module: "Sound - Grade 10 (11) [CAPS]"

Used here as: "Sound"

By: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/m37912/1.5/

Pages: 377-389

Copyright: Mark Horner, Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Sound - Grade 11

By: Rory Adams, Free High School Science Texts Project, Mark Horner, Heather Williams

URL: http://cnx.org/content/m32834/1.2/

Module: "Electromagnetic Radiation - Grade 10 (12) [CAPS]"

Used here as: "Electromagnetic Radiation"

By: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/m37828/1.4/

Pages: 391-401

Copyright: Mark Horner, Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Electromagnetic Radiation - Grade 12

By: Rory Adams, Free High School Science Texts Project, Mark Horner, Heather Williams

URL: http://cnx.org/content/m33652/1.2/

Module: "Magnetism - Grade 10 [CAPS]"

Used here as: "Magnetism"

By: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/m37830/1.5/

Pages: 403-413

Copyright: Mark Horner, Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Magnetism - Grade 10

By: Rory Adams, Free High School Science Texts Project, Mark Horner, Heather Williams

URL: http://cnx.org/content/m32828/1.4/

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Module: "Electrostatics - Grade 10 [CAPS]"

Used here as: "Electrostatics"

By: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/m37825/1.3/

Pages: 415-427

Copyright: Mark Horner, Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Electrostatics - Grade 10

By: Rory Adams, Free High School Science Texts Project, Mark Horner, Heather Williams

URL: http://cnx.org/content/m32829/1.6/

Module: "Electric Circuits - Grade 10 [CAPS]"

Used here as: "Electric Circuits"

By: Mark Horner, Free High School Science Texts Project

URL: http://http://cnx.org/content/m38516/1.2/

Pages: 429-462

Copyright: Mark Horner, Free High School Science Texts Project

License: http://creativecommons.org/licenses/by/3.0/

Based on: Electric Circuits - Grade 10

By: Rory Adams, Free High School Science Texts Project, Mark Horner, Heather Williams

URL: http://cnx.org/content/m32830/1.7/

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